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Chapter 6 – Introduction (6/24/06) Page 6.0-1
CMOS Analog Circuit Design © P.E. Allen - 2006
CHAPTER 6 – CMOS OPERATIONAL AMPLIFIERS INTRODUCTION
Chapter Outline
6.1 CMOS Op Amps 6.2 Compensation of Op Amps 6.3 Two-Stage Operational Amplifier Design 6.4 Cascode Op Amps 6.5 Simulation and Measurement of Op Amps 6.6 Macromodels for Op Amps 6.7 Summary
Hierarchical Perspective:
The op amps in this chapter represent an example of a reasonable complex circuit. The blocks and sub-blocks of the last two chapters will be used to implement the op amp in this chapter.
Blocks or circuits(Combination of primitives, independent)
Sub-blocks or subcircuits(A primitive, not independent)
Functional blocks or circuits(Perform a complex function)
Fig. 6.0-1
Chapter 6
Chapter 6 – Introduction (6/24/06) Page 6.0-2
CMOS Analog Circuit Design © P.E. Allen - 2006
What is an Op Amp?
The op amp (operational amplifier) is a high gain, dc coupled amplifier designed to be used with negative feedback to precisely define a closed loop transfer function.
The basic requirements for an op amp:
• Sufficiently large gain (the accuracy of the signal processing determines this)
• Differential inputs
• Frequency characteristics that permit stable operation when negative feedback is applied
Other requirements:
• High input impedance
• Low output impedance
• High speed/frequency
Chapter 6 – Introduction (6/24/06) Page 6.0-3
CMOS Analog Circuit Design © P.E. Allen - 2006
Why Op Amps?
The op amp is designed to be used with single-loop, negative feedback to accomplish precision signal processing as illustrated below.
060625-01
+−
Σ
F(s)
A(s)
Op Amp
Feedback Network
Vin(s) Vout(s)Av(s)
Vout(s)Vin(s) +
−
F(s)Vf(s) Vf(s)
Single-Loop Negative Feedback Network Op Amp Implementation of a Single-LoopNegative Feedback Network
The voltage gain, Vout(s)Vin(s) , can be shown to be equal to,
Vout(s)Vin(s) =
Av(s)1+Av(s)F(s)
If the product of Av(s)F(s) is much greater than 1, then the voltage gain becomes,
Vout(s)Vin(s)
1F(s) The precision of the voltage gain is defined by F(s).
Chapter 6 – Section 1 (6/24/06) Page 6.1-1
CMOS Analog Circuit Design © P.E. Allen - 2006
SECTION 6.1 –CMOS OPERATIONAL AMPLIFIERS APPLICATION OF THE OP AMP
Ideal Op Amp Symbol:
Null port: If the differential gain of the op amp is large enough then input terminal pair becomes a null port. A null port is a pair of terminals where the voltage is zero and the current is zero. I.e.,
v1 - v2 = vi = 0 and
i1 = 0 and i2 = 0 Therefore, ideal op amps can be analyzed by assuming the differential input voltage is zero and that no current flows into or out of the differential inputs.
+
-
+
-
+
-
v1
v2vOUT = Av(v1-v2)
VDD
VSS
Fig. 110-02
+
-
i1
i2+
-vi
Chapter 6 – Section 1 (6/24/06) Page 6.1-2
CMOS Analog Circuit Design © P.E. Allen - 2006
General Configuration of the Op Amp as a Voltage Amplifier
+-
+
-
+
-
+
-v1
v2 vout
Fig. 110-03
vinpvinn
R1 R2
Non-inverting voltage amplifier:
vinn = 0 vout = R1+R2
R1vinp
Inverting voltage amplifier:
vinp = 0 vout = -R2R1
vinn
Chapter 6 – Section 1 (6/24/06) Page 6.1-3
CMOS Analog Circuit Design © P.E. Allen - 2006
Example 6.1-1 - Simplified Analysis of an Op Amp Circuit
The circuit shown below is an inverting voltage amplifier using an op amp. Find the voltage transfer function, vout/vin.
Solution
If Av , then vi 0 because of the negative feedback path through R2. (The op amp with –fb. makes its input terminal voltages equal.)
vi = 0 and ii = 0 Note that the null port becomes the familiar virtual ground if one of the op amp input terminals is on ground. If this is the case, then we can write that
i1 = vinR1
and i2 = voutR2
Since, ii = 0, then i1 + i2 = 0 giving the desired result as voutvin = -
R2R1 .
+- +
-
+
-
+
-vin vi vout
R2R1
ii
i1 i2
Virtual Ground Fig. 110-04
Chapter 6 – Section 1 (6/24/06) Page 6.1-4
CMOS Analog Circuit Design © P.E. Allen - 2006
Example 6.1-2 – A More Challenging Example of Ideal Op Amp Analysis
Solve for Zin of the op amp circuit shown assuming the op amps are ideal.
Solution
Relationships enforced by the op amps:
1.) Iin = I1 2.) V1 = V2 3.) I2 + I3 = 0 4.) V3 = V4 5.) I4 + I5 = 0 6.) Vin = V5
Therefore,
Iin = I1 = V1Z1 =
V2Z1 =
I2Z2Z1 =
-I3Z2Z1 =
-V3Z2Z1Z3 =
-V4Z2Z1Z3 =
-I4Z2Z4Z1Z3 =
I5Z2Z4Z1Z3
Now, solve for Zin,
Zin = VinIin =
V5Iin =
V5Z1Z3 I5Z2Z4 =
Z1Z3Z5 Z2Z4
060625-02
+ −
Z1 Z2 Z3 Z4
+−Z5
+
−
Zin(s) = Vin
Iin
+ −V1
I1 + −V3
I3+−
V2
I2 +−V4
I4
I5
+
−V5
VinIin
A1
A2
Chapter 6 – Section 1 (6/24/06) Page 6.1-5
CMOS Analog Circuit Design © P.E. Allen - 2006
OP AMP CHARACTERIZATION
Linear and Static Characterization of the CMOS Op Amp A model for a nonideal op amp that includes some of the linear, static nonidealities:
060625-03
+
-v2
v1
v1CMRR
VOS
Ricm
Ricm
en2
Cid RidRout vout
Ideal Op Amp
*
Cicm
Cicm
where
Rid = differential input resistance Cid = differential input capacitance Ricm = common mode input resistance CRicm = common mode input capacitance VOS = input-offset voltage CMRR = common-mode rejection ratio (when v1=v2 an output results)
e2n = voltage-noise spectral density (mean-square volts/Hertz)
Chapter 6 – Section 1 (6/24/06) Page 6.1-6
CMOS Analog Circuit Design © P.E. Allen - 2006
Linear and Dynamic Characteristics of the Op Amp Differential and common-mode frequency response:
Vout(s) = Av(s)[V1(s) - V2(s)] ± Ac(s) V1(s)+V2(s)
2
Differential-frequency response:
Av(s) = Av0
sp1
- 1s
p2 - 1
sp3
- 1 ··· =
Av0 p1p2p3···(s -p1)(s -p2)(s -p3)···
where p1, p2, p3,··· are the poles of the differential-frequency response (ignoring zeros).
0dB
20log10(Av0)
|Av(jω)| dB
AsymptoticMagnitude
ActualMagnitude
ω1
ω2 ω3ω
-6dB/oct.
-12dB/oct.
-18dB/oct.
GB
Fig. 110-06
Chapter 6 – Section 1 (6/24/06) Page 6.1-7
CMOS Analog Circuit Design © P.E. Allen - 2006
Other Characteristics of the Op Amp
Power supply rejection ratio (PSRR):
PSRR = VDDVOUT
Av(s) = Vo/Vin (Vdd = 0)Vo/Vdd (Vin = 0)
Input common mode range (ICMR): ICMR = the voltage range over which the input common-mode signal can vary
without influence the differential performance Slew rate (SR): SR = output voltage rate limit of the op amp Settling time (Ts):
+-
Settling Time
Final Value
Final Value + ε
Final Value - ε
ε
ε
vOUT(t)
t00
vOUTvIN
Fig. 110-07Ts
Upper Tolerance
Lower Tolerance
Chapter 6 – Section 1 (6/24/06) Page 6.1-8
CMOS Analog Circuit Design © P.E. Allen - 2006
OP AMP CATEGORIZATION Classification of CMOS Op Amps
Conversion
Classic DifferentialAmplifier
Modified DifferentialAmplifier
Differential-to-single endedLoad (Current Mirror)
Source/SinkCurrent Loads
MOS DiodeLoad
TransconductanceGrounded Gate
TransconductanceGrounded Source
Class A (Sourceor Sink Load)
Class B(Push-Pull)
Voltageto Current
Currentto Voltage
Voltageto Current
Currentto Voltage
Hierarchy
FirstVoltageStage
SecondVoltageStage
CurrentStage
Table 110-01
Chapter 6 – Section 1 (6/24/06) Page 6.1-9
CMOS Analog Circuit Design © P.E. Allen - 2006
Two-Stage CMOS Op Amp Classical two-stage CMOS op amp broken into voltage-to-current and current-to-voltage stages:
+
--
+
vin
M1 M2
M3 M4
M5
M6
M7
vout
VDD
VSSV→I I→V V→I I→V
voutvin
VBias
Fig. 6.1-8
Chapter 6 – Section 1 (6/24/06) Page 6.1-10
CMOS Analog Circuit Design © P.E. Allen - 2006
Folded Cascode CMOS Op Amp
Folded cascode CMOS op amp broken into stages.
060118-10VSS
VDD
M1 M2
M6
M4
M3
M5
M7
M8
M10
M9
M11
VBias
VBias
VPBias1
+
-vin vout
+
-
V→I I→I I→V
voutvin
VPBias2
Chapter 6 – Section 1 (6/24/06) Page 6.1-11
CMOS Analog Circuit Design © P.E. Allen - 2006
DESIGN OF CMOS OP AMPS Steps in Designing a CMOS Op Amp Steps: 1.) Choosing or creating the basic structure of the op amp. This step is results in a schematic showing the transistors and their interconnections. This diagram does not change throughout the remainder of the design unless the
specifications cannot be met, then a new or modified structure must be developed. 2.) Selection of the dc currents and transistor sizes. Most of the effort of design is in this category. Simulators are used to aid the designer in this phase. 3.) Physical implementation of the design. Layout of the transistors Floorplanning the connections, pin-outs, power supply buses and grounds Extraction of the physical parasitics and re-simulation Verification that the layout is a physical representation of the circuit. 4.) Fabrication 5.) Measurement Verification of the specifications Modification of the design as necessary
Chapter 6 – Section 1 (6/24/06) Page 6.1-12
CMOS Analog Circuit Design © P.E. Allen - 2006
Design Inputs Boundary conditions:
1. Process specification (VT, K', Cox, etc.) 2. Supply voltage and range 3. Supply current and range 4. Operating temperature and range
Requirements: 1. Gain 2. Gain bandwidth 3. Settling time 4. Slew rate 5. Common-mode input range, ICMR 6. Common-mode rejection ratio, CMRR 7. Power-supply rejection ratio, PSRR 8. Output-voltage swing 9. Output resistance
10. Offset 11. Noise 12. Layout area
Chapter 6 – Section 1 (6/24/06) Page 6.1-13
CMOS Analog Circuit Design © P.E. Allen - 2006
Specifications for a Typical Unbuffered CMOS Op Amp
Boundary Conditions Requirement Process Specification See Tables 3.1-1 and 3.1-2 Supply Voltage ±2.5 V ±10% Supply Current 100 μA Temperature Range 0 to 70°C Specifications Value Gain 70 dB Gainbandwidth 5 MHz Settling Time 1 μsec Slew Rate 5 V/μsec Input CMR ±1.5 V CMRR 60 dB PSRR 60 dB Output Swing ±1.5 V Output Resistance N/A, capacitive load only Offset ±10 mV Noise 100nV/ Hz at 1KHz Layout Area 10,000 min. channel length2
Chapter 6 – Section 1 (6/24/06) Page 6.1-14
CMOS Analog Circuit Design © P.E. Allen - 2006
Outputs of Op Amp Design
The basic outputs of design are:
1.) The topology
2.) The dc currents
3.) The W and L values of transistors
4.) The values of components
060625-06
-
+vin
M1 M2
M3 M4
M5
M6
M7
vout
VDD
VSS
VBias
CL
+
-
Cc
W/L ratios
Topology
DC Currents
L
W
Op amp circuit or systems
specifications
Design ofCMOS
Op Amps
Componentvalues
C R
50µA
Chapter 6 – Section 1 (6/24/06) Page 6.1-15
CMOS Analog Circuit Design © P.E. Allen - 2006
Some Practical Thoughts on Op Amp Design
1.) Decide upon a suitable topology. • Experience is a great help • The topology should be the one capable of meeting most of the specifications • Try to avoid “inventing” a new topology but start with an existing topology
2.) Determine the type of compensation needed to meet the specifications. • Consider the load and stability requirements • Use some form of Miller compensation or a self-compensated approach
3.) Design dc currents and device sizes for proper dc, ac, and transient performance. • This begins with hand calculations based upon approximate design equations. • Compensation components are also sized in this step of the procedure. • After each device is sized by hand, a circuit simulator is used to fine tune the
design
Two basic steps of design:
1.) “First-cut” - this step is to use hand calculations to propose a design that has potential of satisfying the specifications. Design robustness is developed in this step.
2.) Optimization - this step uses the computer to refine and optimize the design.
Chapter 6 – Section 2 (6/24/06) Page 6.2-1
CMOS Analog Circuit Design © P.E. Allen - 2006
SECTION 6.2 - COMPENSATION OF OP AMPS INTRODUCTION TO COMPENSATION
Compensation
Objective
Objective of compensation is to achieve stable operation when negative feedback is applied around the op amp.
Types of Compensation
1. Miller - Use of a capacitor feeding back around a high-gain, inverting stage. • Miller capacitor only • Miller capacitor with an unity-gain buffer to block the forward path through the
compensation capacitor. Can eliminate the RHP zero. • Miller with a nulling resistor. Similar to Miller but with an added series resistance
to gain control over the RHP zero.
2. Self compensating - Load capacitor compensates the op amp (later).
3. Feedforward - Bypassing a positive gain amplifier resulting in phase lead. Gain can be less than unity. Because compensation plays such a strong role in design, it is considered before design.
Chapter 6 – Section 2 (6/24/06) Page 6.2-2
CMOS Analog Circuit Design © P.E. Allen - 2006
Single-Loop, Negative Feedback Systems Block diagram: A(s) = differential-mode voltage gain of the
op amp F(s) = feedback transfer function from the
output of op amp back to the input. Definitions: • Open-loop gain = L(s) = -A(s)F(s)
• Closed-loop gain = Vout(s)Vin(s) =
A(s)1+A(s)F(s)
Stability Requirements: The requirements for stability for a single-loop, negative feedback system is, |A(j 0°)F(j 0°)| = |L(j 0°)| < 1
where 0° = 360° is defined as Arg[ A(j 0°)F(j 0°)] = Arg[L(j 0°)] = 0° = 360° Another convenient way to express this requirement is Arg[ A(j 0dB)F(j 0dB)] = Arg[L(j 0dB)] > 0° where 0dB is defined as |A(j 0dB)F(j 0dB)| = |L(j 0dB)| = 1
A(s)
F(s)
Σ-
+Vin(s) Vout(s)
Fig. 120-01
Chapter 6 – Section 2 (6/24/06) Page 6.2-3
CMOS Analog Circuit Design © P.E. Allen - 2006
Illustration of the Stability Requirement using Bode Plots
060625-07
|A(j
ω)F
(jω
)|
0dB
Arg
[-A
(jω
)F(j
ω)]
180
225
270
315
360 ω0dBω
ω
-20dB/decade
-40dB/decade
ΦM
Frequency (rads/sec.) A measure of stability is given by the phase when |A(j )F(j )| = 1. This phase is called phase margin. Phase margin = M = 360° - Arg[-A(j 0dB)F(j 0dB)] = 360° - Arg[L(j 0dB)]
Chapter 6 – Section 2 (6/24/06) Page 6.2-4
CMOS Analog Circuit Design © P.E. Allen - 2006
Why Do We Want Good Stability? Consider the step response of second-order system which closely models the closed-loop gain of the op amp connected in unity gain.
0
0.2
0.4
0.6
0.8
1.0
1.2
1.4
0 5 10 15
45°50°55°
60°65°
70°vout(t)Av0
ωot = ωnt (sec.)Fig. 120-03
+-
A “good” step response is one that quickly reaches its final value. Therefore, we see that phase margin should be at least 45° and preferably 60° or larger. (A rule of thumb for satisfactory stability is that there should be less than three rings.)
Note that good stability is not necessarily the quickest risetime.
Chapter 6 – Section 2 (6/24/06) Page 6.2-5
CMOS Analog Circuit Design © P.E. Allen - 2006
Uncompensated Frequency Response of Two-Stage Op Amps Two-Stage Op Amps:
Fig. 120-04
-
+vin
M1 M2
M3 M4
M5
M6
M7
vout
VDD
VSS
VBias+
-
-
+vin
Q1 Q2
Q3 Q4
Q5
Q6
Q7
vout
VCC
VEE
VBias+
-
Small-Signal Model:
vout
Fig. 120-05
gm1vin2
R1 C1
+
-v1
gm2vin2 gm4v1 R2 C2 gm6v2
+
-v2 R3 C3
+
-
D1, D3 (C1, C3) D2, D4 (C2, C4) D6, D7 (C6, C7)
Note that this model neglects the base-collector and gate-drain capacitances for purposes of simplification.
Chapter 6 – Section 2 (6/24/06) Page 6.2-6
CMOS Analog Circuit Design © P.E. Allen - 2006
Uncompensated Frequency Response of Two-Stage Op Amps - Continued For the MOS two-stage op amp:
R1 1
gm3 ||rds3||rds1 1
gm3 R2 = rds2|| rds4 and R3 = rds6|| rds7
C1 = Cgs3+Cgs4+Cbd1+Cbd3 C2 = Cgs6+Cbd2+Cbd4 and C3 = CL +Cbd6+Cbd7 For the BJT two-stage op amp:
R1 = 1
gm3 ||r 3||r 4||ro1||ro31
gm3 R2 = r 6|| ro2|| ro4 r 6 and R3 = ro6|| ro7
C1 = C 3+C 4+Ccs1+Ccs3 C2 = C 6+Ccs2+Ccs4 and C3 = CL+Ccs6+Ccs7 Assuming the pole due to C1 is much greater than the poles due to C2 and C3 gives,
voutgm1vinR2 C2 gm6v2
+
-v2 R3 C3
+
-
Fig. 120-06
Voutgm1VinRI CI gmIIVI
+
-VI RII CII
+
-
The locations for the two poles are given by the following equations
p’1 = 1
RICI and p’2 = 1
RIICII
where RI (RII) is the resistance to ground seen from the output of the first (second) stage and CI (CII) is the capacitance to ground seen from the output of the first (second) stage.
Chapter 6 – Section 2 (6/24/06) Page 6.2-7
CMOS Analog Circuit Design © P.E. Allen - 2006
Uncompensated Frequency Response of an Op Amp (F(s) = 1)
060625-08
0dB
Avd(0) dB
-20dB/decade
log10(ω)
log10(ω)
180°
270°
360°
Phase Shift
GB
|p1'|
-40dB/decade
315°
225°
-45/decade
-45/decade
|p2'|
|A(j
ω)|
Arg
[-A
(jω
)]
ω0dB If we assume that F(s) = 1 (this is the worst case for stability considerations), then the above plot is the same as the loop gain. Note that the phase margin is much less than 45° ( 6°). Therefore, the op amp must be compensated before using it in a closed-loop configuration.
Chapter 6 – Section 2 (6/24/06) Page 6.2-8
CMOS Analog Circuit Design © P.E. Allen - 2006
MILLER COMPENSATION
Miller Compensation of the Two-Stage Op Amp
Fig. 120-08
-
+vin
M1 M2
M3 M4
M5
M6
M7
vout
VDD
VSS
VBias+
-
-
+vin
Q1 Q2
Q3 Q4
Q5
Q6
Q7
VCC
VEE
VBias+
-
CM
CI
Cc
CII
vout
CI
Cc
CII
CM
The various capacitors are: Cc = accomplishes the Miller compensation CM = capacitance associated with the first-stage mirror (mirror pole) CI = output capacitance to ground of the first-stage CII = output capacitance to ground of the second-stage
Chapter 6 – Section 2 (6/24/06) Page 6.2-9
CMOS Analog Circuit Design © P.E. Allen - 2006
Compensated Two-Stage, Small-Signal Frequency Response Model Simplified
Use the CMOS op amp to illustrate: 1.) Assume that gm3 >> gds3 + gds1
2.) Assume that gm3CM
>> GB
Therefore,
-gm1vin2 CM
1gm3 gm4v1
gm2vin2 C1 rds2||rds4
gm6v2 rds6||rds7 CL
v1 v2Cc
+
-
vout
Fig. 120-09
rds1||rds3
gm1vin rds2||rds4 gm6v2 rds6||rds7CII
v2Cc
+
-
voutCI
+
-vin
Same circuit holds for the BJT op amp with different component relationships.
Chapter 6 – Section 2 (6/24/06) Page 6.2-10
CMOS Analog Circuit Design © P.E. Allen - 2006
General Two-Stage Frequency Response Analysis where gmI = gm1 = gm2, RI = rds2||rds4, CI = C1 and gmII = gm6, RII = rds6||rds7, CII = C2 = CL Nodal Equations:
-gmIVin = [GI + s(CI + Cc)]V2 - [sCc]Vout and 0 = [gmII - sCc]V2 + [GII + sCII + sCc]Vout Solving using Cramer’s rule gives,
Vout(s)Vin(s) =
gmI(gmII - sCc)GIGII+s [GII(CI+CII)+GI(CII+Cc)+gmIICc]+s2[CICII+CcCI+CcCII]
= Ao[1 - s (Cc/gmII)]
1+s [RI(CI+CII)+RII(C2+Cc)+gmIIR1RIICc]+s2[RIRII(CICII+CcCI+CcCII)]
where, Ao = gmIgmIIRIRII
In general, D(s) = 1-sp1
1-sp2
= 1-s 1p1
+ 1p2
+s2
p1p2 D(s) 1-
sp1
+ s2
p1p2 , if |p2|>>|p1|
p1 = -1
RI(CI+CII)+RII(CII+Cc)+gmIIR1RIICc
-1gmIIR1RIICc
, z = gmII
Cc
gmIVin RI gmIIV2 RII CII
V2Cc
+
-VoutCI
+
-Vin
Fig.120-10
p2 = -[RI(CI+CII)+RII(CII+Cc)+gmIIR1RIICc]
RIRII(CICII+CcCI+CcCII)-gmIICc
CICII+CcCI+CcCII
-gmIICII
, CII > Cc > CI
Chapter 6 – Section 2 (6/24/06) Page 6.2-11
CMOS Analog Circuit Design © P.E. Allen - 2006
Summary of Results for Miller Compensation of the Two-Stage Op Amp
There are three roots of importance:
1.) Right-half plane zero:
z1= gmIICc
= gm6Cc
This root is very undesirable- it boosts the magnitude while decreasing the phase.
2.) Dominant left-half plane pole (the Miller pole):
p1 -1
gmIIRIRIICc =
-(gds2+gds4)(gds6+gds7)gm6Cc
This root accomplishes the desired compensation.
3.) Left-half plane output pole:
p2 -gmIICII
-gm6CL
This pole must be unity-gainbandwidth or the phase margin will not be satisfied. Root locus plot of the Miller compensation:
jω
σ
Cc=0Open-loop poles
Closed-loop poles, Cc≠0
p2 p2' p1p1' z1 Fig. 120-11
Chapter 6 – Section 2 (6/24/06) Page 6.2-12
CMOS Analog Circuit Design © P.E. Allen - 2006
Compensated Open-Loop Frequency Response of the Two-Stage Op Amp
060118-10
0dB
Avd(0) dB
-20dB/decade
log10(ω)
log10(ω)
Phase Margin
180°
270°
360°
Phase Shift
GB
-40dB/decade
315°
225°
|p1'|
No phase margin
Uncompensated
Compensated
-45/decade
-45/decade
|p2'||p1| |p2|
|A(j
ω)F
(jω
)|A
rg[-A
(jω
)F(j
ω)|
Compensated
Uncompensated
F(jω)=1
F(jω)=1
Note that the unity-gainbandwidth, GB, is
GB = Avd(0)·|p1| = (gmIgmIIRIRII)1
gmIIRIRIICc =
gmICc
= gm1Cc
= gm2Cc
Chapter 6 – Section 2 (6/24/06) Page 6.2-13
CMOS Analog Circuit Design © P.E. Allen - 2006
Conceptually, where do these roots come from? 1.) The Miller pole:
|p1| 1
RI(gm6RIICc)
2.) The left-half plane output pole:
|p2| gm6CII
3.) Right-half plane zero (One source of zeros is from multiple paths from the input to output):
vout = -gm6RII(1/sCc)
RII + 1/sCc v’ +
RII
RII + 1/sCc v’’ =
-RII
gm6
sCc - 1
RII + 1/sCc v
where v = v’ = v’’.
VDD
CcRII
vout
vI
M6RI
≈gm6RIICcFig. 120-13
VDD
CcRII
vout
M6CII
GB·Cc
1 ≈ 0
VDD
RII
vout
M6CII
Fig. 120-14 VDD
CcRII
vout
v'v''
M6
Fig. 120-15
Chapter 6 – Section 2 (6/24/06) Page 6.2-14
CMOS Analog Circuit Design © P.E. Allen - 2006
Further Comments on p2 The previous observations on p2 can be proved as follows: Find the resistance RCc seen by the compensation capacitor, Cc.
060626-02
VDD
RII
RI
M6
RCc
RIRII
+
−vgs6gm6vgs6
vx
ix ixRCc
Cc
vx = ixRI + (ix + gm6vgs6)RII = ixRI + (ix + gm6ixRI)RII
Therefore,
RCc = vxix = RI + (1 + gm6RI)RII gm6RIRII
The frequency at which Cc begins to become a short is,
1Cc < gm6RIRII or >
1gm6RIRII Cc |p1|
Thus, at the frequency where CII begins to short the output, Cc is acting as a short.
Chapter 6 – Section 2 (6/24/06) Page 6.2-15
CMOS Analog Circuit Design © P.E. Allen - 2006
Influence of the Mirror Pole Up to this point, we have neglected the influence of the pole, p3, associated with the current mirror of the input stage. A small-signal model for the input stage that includes C3 is shown below:
gm3rds31
rds1
gm1Vin
rds2
i3
i3 rds4C3
+
-Vo1
2gm2Vin
2
Fig. 120-16 The transfer function from the input to the output voltage of the first stage, Vo1(s), can be written as
Vo1(s) Vin(s) =
-gm12(gds2+gds4)
gm3+gds1+gds3gm3+ gds1+gds3+sC3 + 1
-gm12(gds2+gds4)
sC3 + 2gm3 sC3 + gm3
We see that there is a pole and a zero given as
p3 = - gm3C3
and z3 = - 2gm3C3
Chapter 6 – Section 2 (6/24/06) Page 6.2-16
CMOS Analog Circuit Design © P.E. Allen - 2006
Summary of the Conditions for Stability of the Two-Stage Op Amp
• Unity-gainbandwith is given as:
GB = Av(0)·|p1| =(gmIgmIIRIRII)·1
gmIIRIRIICc =
gmICc
= (gm1gm2R1R2)·1
gm2R1R2Cc =
gm1Cc
• The requirement for 45° phase margin is:
±180° - Arg[Loop Gain] = ±180° - tan-1 |p1| - tan-1 |p2| - tan-1 z = 45°
Let = GB and assume that z 10GB, therefore we get,
±180° - tan-1GB|p1| - tan-1
GB|p2| - tan-1
GBz = 45°
135° tan-1(Av(0)) + tan-1GB|p2| + tan-1(0.1) = 90° + tan-1
GB|p2| + 5.7°
39.3° tan-1GB|p2|
GB|p2| = 0.818 |p2| 1.22GB
• The requirement for 60° phase margin: |p2| 2.2GB if z 10GB
• If 60° phase margin is required, then the following relationships apply:
gm6Cc >
10gm1Cc gm6 > 10gm1 and
gm6C2 >
2.2gm1Cc Cc > 0.22C2
Chapter 6 – Section 2 (6/24/06) Page 6.2-17
CMOS Analog Circuit Design © P.E. Allen - 2006
RIGHT-HALF PLANE ZERO
Controlling the Right-Half Plane Zero Why is the RHP zero a problem? Because it boosts the magnitude but lags the phase - the worst possible combination for stability.
060626-03
jω
σ
jω1
jω2
jω3
θ1θ2
θ3
180 > θ1 > θ2 > θ3
z1
LoopGain
0dB log10ω
log10ω360°
180°
RHP Zero Boost
RHP Zero Lag
Loop PhaseShift
Solution of the problem:
The compensation comes from the feedback path through Cc, but the RHP zero
comes from the feedforward path through Cc so eliminate the feedforward path!
Chapter 6 – Section 2 (6/24/06) Page 6.2-18
CMOS Analog Circuit Design © P.E. Allen - 2006
Use of Buffer to Eliminate the Feedforward Path through the Miller Capacitor Model: The transfer function is given by the following equation,
Vo(s)Vin(s) =
(gmI)(gmII)(RI)(RII)1 + s[RICI + RIICII + RICc + gmIIRIRIICc] + s2[RIRIICII(CI + Cc)]
Using the technique as before to approximate p1 and p2 results in the following
p1 1
RICI + RIICII + RICc + gmIIRIRIICc 1
gmIIRIRIICc
and
p2 gmIICc
CII(CI + Cc)
Comments: Poles are approximately what they were before with the zero removed. For 45° phase margin, |p2| must be greater than GB For 60° phase margin, |p2| must be greater than 1.73GB
Fig. 430-02
InvertingHigh-GainStage
Cc
vOUT gmIvin RIgmIIVI
RII CII
VICc
+
-VoutCI
+
-Vin Vout
+1
Chapter 6 – Section 2 (6/24/06) Page 6.2-19
CMOS Analog Circuit Design © P.E. Allen - 2006
Use of Buffer with Finite Output Resistance to Eliminate the RHP Zero Assume that the unity-gain buffer has an output resistance of Ro. Model:
InvertingHigh-GainStage
+1Cc
vOUT gmIvin RIgmIIVI
RII CII
VICc
+
-
VoutCI
+
-Vin Ro
Ro
Vout
Fig. 430-03
Ro
It can be shown that if the output resistance of the buffer amplifier, Ro, is not neglected that another pole occurs at,
p4 1
Ro[CICc/(CI + Cc)]
and a LHP zero at
z2 1
RoCc
Closer examination shows that if a resistor, called a nulling resistor, is placed in series with Cc that the RHP zero can be eliminated or moved to the LHP.
Chapter 6 – Section 2 (6/24/06) Page 6.2-20
CMOS Analog Circuit Design © P.E. Allen - 2006
Use of Nulling Resistor to Eliminate the RHP Zero (or turn it into a LHP zero)†
InvertingHigh-GainStage
Cc
vOUT
Rz
gmIvin RIgmIIVI RII CII
Cc
+
-
VoutCI
+
-Vin
Rz
Fig. 430-04
VI
Nodal equations:
gmIVin + VIRI
+ sCIVI + sCc
1 + sCcRz (VI Vout) = 0
gmIIVI + VoRII
+ sCIIVout + sCc
1 + sCcRz (Vout VI) = 0
Solution:
Vout(s)Vin(s) =
a{1 s[(Cc/gmII) RzCc]}1 + bs + cs2 + ds3
where
† W,J. Parrish, "An Ion Implanted CMOS Amplifier for High Performance Active Filters", Ph.D. Dissertation, 1976, Univ. of CA., Santa Barbara.
a = gmIgmIIRIRII b = (CII + Cc)RII + (CI + Cc)RI + gmIIRIRIICc + RzCc c = [RIRII(CICII + CcCI + CcCII) + RzCc(RICI + RIICII)] d = RIRIIRzCICIICc
Chapter 6 – Section 2 (6/24/06) Page 6.2-21
CMOS Analog Circuit Design © P.E. Allen - 2006
Use of Nulling Resistor to Eliminate the RHP - Continued If Rz is assumed to be less than RI or RII and the poles widely spaced, then the roots of the above transfer function can be approximated as
p1 1
(1 + gmIIRII)RICc 1
gmIIRIIRICc
p2 gmIICc
CICII + CcCI + CcCII gmIICII
p4 = 1
RzCI
and
z1 = 1
Cc(1/gmII Rz)
Note that the zero can be placed anywhere on the real axis.
Chapter 6 – Section 2 (6/24/06) Page 6.2-22
CMOS Analog Circuit Design © P.E. Allen - 2006
Conceptual Illustration of the Nulling Resistor Approach VDD
CcRII
Vout
V'V''
M6
Rz
Fig. Fig. 430-05 The output voltage, Vout, can be written as
Vout = -gm6RII Rz +
1sCc
RII + Rz + 1
sCc
V’ + RII
RII + Rz + 1
sCc
V” = -RII gm6Rz +
gm6sCc
- 1
RII + Rz + 1
sCc
V
when V = V’ = V’’. Setting the numerator equal to zero and assuming gm6 = gmII gives,
z1 = 1
Cc(1/gmII Rz)
Chapter 6 – Section 2 (6/24/06) Page 6.2-23
CMOS Analog Circuit Design © P.E. Allen - 2006
A Design Procedure that Allows the RHP Zero to Cancel the Output Pole, p2 We desire that z1 = p2 in terms of the previous notation. Therefore,
1
Cc(1/gmII Rz) = gmIICII
The value of Rz can be found as
Rz = Cc + CII
Cc (1/gmII)
With p2 canceled, the remaining roots are p1 and p4(the pole due to Rz) . For unity-gain stability, all that is required is that
|p4| > Av(0)|p1| = Av(0)
gmIIRIIRICc = gmICc
and (1/RzCI) > (gmI/Cc) = GB
Substituting Rz into the above inequality and assuming CII >> Cc results in
Cc > gmIgmII CICII
This procedure gives excellent stability for a fixed value of CII ( CL). Unfortunately, as CL changes, p2 changes and the zero must be readjusted to cancel p2.
jω
Fig. 430-06σ
-p4 -p2 -p1 z1
Chapter 6 – Section 2 (6/24/06) Page 6.2-24
CMOS Analog Circuit Design © P.E. Allen - 2006
Increasing the Magnitude of the Output Pole† The magnitude of the output pole , p2, can be increased by introducing gain in the Miller capacitor feedback path. For example,
VDD
VSS
VBias
Cc
M6
M7
M8
M9M10
M12M11
vOUTIin R1 R2 C2
rds8
gm8Vs8
Cc
VoutV1
+
-
+
-
+
-Vs8
Iin R1 R2 C2gm8Vs8
VoutV1
+
-
+
-
+
-Vs8
1gm8
Cc
gm6V1
gm6V1
Cgd6
Cgd6
Fig. 6.2-15B The resistors R1 and R2 are defined as
R1 = 1
gds2 + gds4 + gds9 and R2 =
1gds6 + gds7
where transistors M2 and M4 are the output transistors of the first stage. Nodal equations:
† B.K. Ahuja, “An Improved Frequency Compensation Technique for CMOS Operational Amplifiers,” IEEE J. of Solid-State Circuits, Vol. SC-18,
No. 6 (Dec. 1983) pp. 629-633.
Iin = G1V1-gm8Vs8 = G1V1-gm8sCc
gm8 + sCc Vout and 0 = gm6V1+ G2+sC2+
gm8sCcgm8+sCc
Vout
Chapter 6 – Section 2 (6/24/06) Page 6.2-25
CMOS Analog Circuit Design © P.E. Allen - 2006
Increasing the Magnitude of the Output Pole - Continued Solving for the transfer function Vout/Iin gives,
VoutIin
= -gm6G1G2
1 +
sCcgm8
1 + s Ccgm8
+ C2G2
+ CcG2
+ gm6CcG1G2
+ s2 CcC2
gm8G2
Using the approximate method of solving for the roots of the denominator gives
p1 = -1
Ccgm8
+ CcG2
+ C2G2
+ gm6CcG1G2
-6
gm6rds2Cc
and p2 -
gm6rds2Cc6
CcC2gm8G2
= gm8rds2G2
6 gm6C2
= gm8rds
3 |p2’|
where all the various channel resistance have been assumed to equal rds and p2’ is the output pole for normal Miller compensation. Result: Dominant pole is approximately the same and the output pole is increased by gmrds.
Chapter 6 – Section 2 (6/24/06) Page 6.2-26
CMOS Analog Circuit Design © P.E. Allen - 2006
Increasing the Magnitude of the Output Pole - Continued In addition there is a LHP zero at -gm8/sCc and a RHP zero due to Cgd6 (shown dashed in the model on Page 6.2-24) at gm6/Cgd6. Roots are:
jω
σgm6Cgd6
-gm8Cc
-gm6gm8rds3C2
-1gm6rdsCc Fig. 6.2-16A
Chapter 6 – Section 2 (6/24/06) Page 6.2-27
CMOS Analog Circuit Design © P.E. Allen - 2006
Concept Behind the Increasing of the Magnitude of the Output Pole
VDD
Ccrds7
vout
M6 CII
GB·Cc
1 ≈ 0
VDD
vout
M6CII
M8
gm8rds8
Fig. Fig. 430-08
rds73
Rout = rds7||3
gm6gm8rds8 3
gm6gm8rds8
Therefore, the output pole is approximately,
|p2| gm6gm8rds8
3CII
Chapter 6 – Section 2 (6/24/06) Page 6.2-28
CMOS Analog Circuit Design © P.E. Allen - 2006
FINDING ROOTS BY INSPECTION Identification of Poles from a Schematic
1.) Most poles are equal to the reciprocal product of the resistance from a node to ground and the capacitance connected to that node.
2.) Exceptions (generally due to feedback):
a.) Negative feedback:
-A
R1
C2
C1
C3
-A
R1
C2
C1 C3(1+A)RootID01
b.) Positive feedback (A<1):
+A
R1
C2
C1
C3
+A
R1
C2
C1 C3(1-A)RootID02
Chapter 6 – Section 2 (6/24/06) Page 6.2-29
CMOS Analog Circuit Design © P.E. Allen - 2006
Identification of Zeros from a Schematic 1.) Zeros arise from poles in
the feedback path.
If F(s) = 1
sp1
+1 , then
VoutVin
= A(s)
1+A(s)F(s) = A(s)
1+A(s)1
sp1
+1
=A(s)
sp1
+1
sp1
+1+ A(s)
2.) Zeros are also created by two paths from the input to the output and one of
more of the paths is frequency dependent.
vin vout
F(s)
A(s)Σ−
+RootID03
VDD
CcRII
vout
v'v''
M6
Fig. 120-15
Chapter 6 – Section 2 (6/24/06) Page 6.2-30
CMOS Analog Circuit Design © P.E. Allen - 2006
OTHER FORMS OF COMPENSATION Feedforward Compensation Use two parallel paths to achieve a LHP zero for lead compensation purposes.
CcA
VoutVi
InvertingHigh GainAmplifier
CII RII
RHP Zero Cc-A
VoutVi
InvertingHigh GainAmplifier
CII RII
LHP Zero
A
CII RIIVi Vout
Cc
gmIIVi
+
-
+
- Fig.430-09
Cc
VoutVi +1
LHP Zero using Follower
Vout(s)Vin(s) =
ACcCc + CII
s + gmII/ACc
s + 1/[RII(Cc + CII)]
To use the LHP zero for compensation, a compromise must be observed. • Placing the zero below GB will lead to boosting of the loop gain that could deteriorate
the phase margin. • Placing the zero above GB will have less influence on the leading phase caused by the
zero. Note that a source follower is a good candidate for the use of feedforward compensation.
Chapter 6 – Section 2 (6/24/06) Page 6.2-31
CMOS Analog Circuit Design © P.E. Allen - 2006
Self-Compensated Op Amps
Self compensation occurs when the load capacitor is the compensation capacitor (can never be unstable for resistive feedback)
Fig. 430-10
-
+vin vout
CL
+
-Gm
Rout(must be large)
Increasing CL
|dB|
Av(0) dB
0dB ω
Rout
-20dB/dec.
Voltage gain:
voutvin
= Av(0) = GmRout
Dominant pole:
p1 = -1
RoutCL
Unity-gainbandwidth:
GB = Av(0)·|p1| = GmCL
Stability: Large load capacitors simply reduce GB but the phase is still 90° at GB.
Chapter 6 – Section 2 (6/24/06) Page 6.2-32
CMOS Analog Circuit Design © P.E. Allen - 2006
CMOS OP AMP SLEW RATE Slew Rate of a Two-Stage CMOS Op Amp Remember that slew rate occurs when currents flowing in a capacitor become limited and is given as
Ilim = C dvCdt where vC is the voltage across the capacitor C.
-
+vin>>0
M1 M2
M3 M4
M5
M6
M7
vout
VDD
VSS
VBias+
-
Cc
CL
I5
Assume a virtural ground
I7
I6I5 ICL
Positive Slew Rate
-
+vin<<0
M1 M2
M3 M4
M5
M6
M7
vout
VDD
VSS
VBias+
-
Cc
CL
I5
Assume a virtural ground
I7
I6=0I5 ICL
Negative Slew Rate Fig. 140-05
SR+ = minI5Cc
, I6-I5-I7
CL =
I5Cc
because I6>>I5 SR- = minI5Cc
, I7-I5CL
= I5Cc
if I7>>I5.
Therefore, if CL is not too large and if I7 is significantly greater than I5, then the slew rate of the two-stage op amp should be, I5/Cc.
Chapter 6 – Section 3 (6/24/06) Page 6.3-1
CMOS Analog Circuit Design © P.E. Allen - 2006
SECTION 6.3 - TWO-STAGE OP AMP DESIGN A DESIGN PROCEDURE FOR THE TWO-STAGE CMOS OP AMP
Unbuffered, Two-Stage CMOS Op Amp
-
+
vin
M1 M2
M3 M4
M5
M6
M7
vout
VDD
VSS
VBias+
-
Cc
CL
Fig. 6.3-1 Notation:
Si = WiLi
= W/L of the ith transistor
Chapter 6 – Section 3 (6/24/06) Page 6.3-2
CMOS Analog Circuit Design © P.E. Allen - 2006
DC Balance Conditions for the Two-Stage Op Amp For best performance, keep all transistors in saturation. M4 is the only transistor that cannot be forced into saturation by internal connections or external voltages. Therefore, we develop conditions to force M4 to be in saturation. 1.) First assume that VSG4 = VSG6. This will cause “proper mirroring” in the M3-M4 mirror. Also, the gate and drain of M4 are at the same potential so that M4 is “guaranteed” to be in saturation.
-
+vin
M1 M2
M3 M4
M5
M6
M7
vout
VDD
VSS
VBias+
-
Cc
CL
-+VSG6
-+VSG4
I4
I5
I7
I6
Fig. 6.3-1A
2.) If VSG4 = VSG6, then I6 = S6S4
I4
3.) However, I7 = S7S5
I5 = S7S5
(2I4)
4.) For balance, I6 must equal I7 S6S4
= 2S7S5
called the “balance conditions”
5.) So if the balance conditions are satisfied, then VDG4 = 0 and M4 is saturated.
Chapter 6 – Section 3 (6/24/06) Page 6.3-3
CMOS Analog Circuit Design © P.E. Allen - 2006
Design Relationships for the Two-Stage Op Amp
Slew rate SR = I5Cc (Assuming I7 >>I5 and CL > Cc)
First-stage gain Av1 = gm1
gds2 + gds4 = 2gm1
I5( 2 + 4)
Second-stage gain Av2 = gm6
gds6 + gds7 = gm6
I6( 6 + 7)
Gain-bandwidth GB = gm1Cc
Output pole p2 = gm6CL
RHP zero z1 = gm6Cc
60° phase margin requires that gm6 = 2.2gm2(CL/Cc) if all other roots are 10GB.
Positive ICMR Vin(max) = VDD I53 |VT03|(max) + VT1(min))
Negative ICMR Vin(min) = VSS + I51 + VT1(max) + VDS5(sat)
Chapter 6 – Section 3 (6/24/06) Page 6.3-4
CMOS Analog Circuit Design © P.E. Allen - 2006
Op Amp Specifications
The following design procedure assumes that specifications for the following parameters are given.
1. Gain at dc, Av(0) 2. Gain-bandwidth, GB 3. Phase margin (or settling time) 4. Input common-mode range, ICMR 5. Load Capacitance, CL 6. Slew-rate, SR 7. Output voltage swing 8. Power dissipation, Pdiss
-
+
vin M1 M2
M3 M4
M5
M6
M7
vout
VDD
VSS
VBias+
-
Cc
CL
VSG4+
-
Max. ICMRand/or p3
VSG6+
-
Vout(max)
I6
gm6 or Proper Mirroring
VSG4=VSG6
Cc ≈ 0.2CL(PM = 60°)
GB =gm1Cc
Min. ICMR I5 I5 = SR·Cc Vout(min)
Fig. 160-02
Chapter 6 – Section 3 (6/24/06) Page 6.3-5
CMOS Analog Circuit Design © P.E. Allen - 2006
Unbuffered Op Amp Design Procedure This design procedure assumes that the gain at dc (Av), unity gain bandwidth (GB), input common mode range (Vin(min) and Vin(max)), load capacitance (CL), slew rate (SR), settling time (Ts), output voltage swing (Vout(max) and Vout(min)), and power dissipation (Pdiss) are given. Choose the smallest device length which will keep the channel modulation parameter constant and give good matching for current mirrors. 1. From the desired phase margin, choose the minimum value for Cc, i.e. for a 60° phase
margin we use the following relationship. This assumes that z 10GB. Cc > 0.22CL 2. Determine the minimum value for the “tail current” (I5) from the largest of the two
values. I5 = SR .Cc or I5 10
VDD + |VSS|
2 .Ts
3. Design for S3 from the maximum input voltage specification.
S3 = I5
K'3[VDD Vin(max) |VT03|(max) + VT1(min)]2
4. Verify that the pole of M3 due to Cgs3 and Cgs4 (= 0.67W3L3Cox) will not be dominant by assuming it to be greater than 10 GB
gm3
2Cgs3 > 10GB.
Chapter 6 – Section 3 (6/24/06) Page 6.3-6
CMOS Analog Circuit Design © P.E. Allen - 2006
Unbuffered Op Amp Design Procedure - Continued 5. Design for S1 (S2) to achieve the desired GB.
gm1 = GB . Cc S2 = gm12
K'1I5
6. Design for S5 from the minimum input voltage. First calculate VDS5(sat) then find S5.
VDS5(sat) = Vin(min) VSS I5
1 VT1(max) 100 mV S5 = 2I5
K'5[VDS5(sat)]2
7. Find S6 by letting the second pole (p2) be equal to 2.2 times GB and assuming that VSG4 = VSG6.
gm6 = 2.2gm2(CL/Cc) and gm6gm4 =
2KP'S6I6
2KP'S4I4 =
S6S4
I6I4
= S6S4
S6 = gm6gm4S4
8. Calculate I6 from
I6 = gm62
2K'6S6
Check to make sure that S6 satisfies the Vout(max) requirement and adjust as necessary. 9. Design S7 to achieve the desired current ratios between I5 and I6. S7 = (I6/I5)S5 (Check the minimum output voltage requirements)
Chapter 6 – Section 3 (6/24/06) Page 6.3-7
CMOS Analog Circuit Design © P.E. Allen - 2006
Unbuffered Op Amp Design Procedure - Continued
10. Check gain and power dissipation specifications.
Av = 2gm2gm6
I5( 2 + 4)I6( 6 + 7) Pdiss = (I5 + I6)(VDD + |VSS|)
11. If the gain specification is not met, then the currents, I5 and I6, can be decreased or the W/L ratios of M2 and/or M6 increased. The previous calculations must be rechecked to insure that they are satisfied. If the power dissipation is too high, then one can only reduce the currents I5 and I6. Reduction of currents will probably necessitate increase of some of the W/L ratios in order to satisfy input and output swings.
12. Simulate the circuit to check to see that all specifications are met.
Chapter 6 – Section 3 (6/24/06) Page 6.3-8
CMOS Analog Circuit Design © P.E. Allen - 2006
DESIGN EXAMPLE
Example 6.3-1 - Design of a Two-Stage Op Amp Using the material and device parameters given in Tables 3.1-1 and 3.1-2, design an amplifier similar to that shown in Fig. 6.3-1 that meets the following specifications. Assume the channel length is to be 1μm and the load capacitor is CL = 10pF. Av > 3000V/V VDD = 2.5V VSS = -2.5V GB = 5MHz SR > 10V/μs 60° phase margin Vout range = ±2V ICMR = -1 to 2V Pdiss 2mW Solution 1.) The first step is to calculate the minimum value of the compensation capacitor Cc, Cc > (2.2/10)(10 pF) = 2.2 pF 2.) Choose Cc as 3pF. Using the slew-rate specification and Cc calculate I5. I5 = (3x10-12)(10x106) = 30 μA 3.) Next calculate (W/L)3 using ICMR requirements.
(W/L)3 = 30x10-6
(50x10-6)[2.5 2 .85 + 0.55]2 = 15 (W/L)3 = (W/L)4 = 15
Chapter 6 – Section 3 (6/24/06) Page 6.3-9
CMOS Analog Circuit Design © P.E. Allen - 2006
Example 6.3-1 - Continued
4.) Now we can check the value of the mirror pole, p3, to make sure that it is in fact greater than 10GB. Assume the Cox = 0.4fF/μm2. The mirror pole can be found as
p3 -gm32Cgs3
= - 2K’pS3I3
2(0.667)W3L3Cox = -2.81x109(rads/sec)
or 448 MHz. Thus, p3, is not of concern in this design because p3 >> 10GB. 5.) The next step in the design is to calculate gm1 to get gm1 = (5x106)(2 )(3x10-12) = 94.25μS Therefore, (W/L)1 is
(W/L)1 = (W/L)2 = gm12
2K’NI1 =
(94.25)2
2·110·15 = 2.79 3.0 (W/L)1 = (W/L)2 = 3
6.) Next calculate VDS5,
VDS5 = ( 1) ( 2.5) 30x10-6
110x10-6·3 - .85 = 0.35V
Using VDS5 calculate (W/L)5 from the saturation relationship.
(W/L)5 = 2(30x10-6)
(110x10-6)(0.35)2 = 4.49 4.5 (W/L)5 = 4.5
Chapter 6 – Section 3 (6/24/06) Page 6.3-10
CMOS Analog Circuit Design © P.E. Allen - 2006
Example 6.3-1 - Continued
7.) For 60° phase margin, we know that gm6 10gm1 942.5μS Assuming that gm6 = 942.5μS and knowing that gm4 = 150μS, we calculate (W/L)6 as
(W/L)6 = 15 942.5x10-6(150x10-6) = 94.25 94
8.) Calculate I6 using the small-signal gm expression:
I6 = (942.5x10-6)2
(2)(50x10-6)(94.25) = 94.5μA 95μA
If we calculate (W/L)6 based on Vout(max), the value is approximately 15. Since 94 exceeds the specification and maintains better phase margin, we will stay with (W/L)6 = 94 and I6 = 95μA. With I6 = 95μA the power dissipation is
Pdiss = 5V·(30μA+95μA) = 0.625mW.
Chapter 6 – Section 3 (6/24/06) Page 6.3-11
CMOS Analog Circuit Design © P.E. Allen - 2006
Example 6.3-1 - Continued
9.) Finally, calculate (W/L)7
(W/L)7 = 4.5 95x10-630x10-6 = 14.25 14 (W/L)7 = 14
Let us check the Vout(min) specification although the W/L of M7 is so large that this is probably not necessary. The value of Vout(min) is
Vout(min) = VDS7(sat) = 2·95
110·14 = 0.351V
which is less than required. At this point, the first-cut design is complete. 10.) Now check to see that the gain specification has been met
Av = (92.45x10-6)(942.5x10-6)
15x10-6(.04 + .05)95x10-6(.04 + .05) = 7,697V/V
which exceeds the specifications by a factor of two. .An easy way to achieve more gain would be to increase the W and L values by a factor of two which because of the decreased value of would multiply the above gain by a factor of 20.
11.) The final step in the hand design is to establish true electrical widths and lengths based upon L and W variations. In this example L will be due to lateral diffusion only. Unless otherwise noted, W will not be taken into account. All dimensions will be rounded to integer values. Assume that L = 0.2μm.
Chapter 6 – Section 3 (6/24/06) Page 6.3-12
CMOS Analog Circuit Design © P.E. Allen - 2006
Example 6.3-1 – Continued
Therefore, we have, W1 = W2 = 3(1 0.4) = 1.8 μm 2μm W3 = W4 = 15(1 0.4) = 9μm W5 = 4.5(1 - 0.4) = 2.7μm 3μm W6 = 94(1 - 0.4) = 56.4μm 56μm W7 = 14(1 - 0.4) = 8.4 8μm The figure below shows the results of the first-cut design. The W/L ratios shown do not account for the lateral diffusion discussed above. The next phase requires simulation.
-
+
vin
M1 M2
M3 M4
M5
M6
M7
vout
VDD = 2.5V
VSS = -2.5V
Cc = 3pF
CL =10pF
3μm1μm
3μm1μm
15μm1μm
15μm1μm
M84.5μm1μm
30μA
4.5μm1μm
14μm1μm
94μm1μm
30μA
95μA
Fig. 6.3-3
Chapter 6 – Section 3 (6/24/06) Page 6.3-13
CMOS Analog Circuit Design © P.E. Allen - 2006
ELIMINATING THE RHP ZERO Incorporating the Nulling Resistor into the Miller Compensated Two-Stage Op Amp Circuit:
We saw earlier that the roots were:
p1 = gm2
AvCc = gm1
AvCc p2 = gm6CL
p4 = 1
RzCI z1 = 1
RzCc Cc/gm6
where Av = gm1gm6RIRII. (Note that p4 is the pole resulting from the nulling resistor compensation technique.)
VDD
VSS
IBias
CL
CcCM vout
VBVA
M1 M2
M3 M4
M5
M6
M7M9
M10
M11
M12
vin+vin-
M8
Fig. 160-03
VC
Chapter 6 – Section 3 (6/24/06) Page 6.3-14
CMOS Analog Circuit Design © P.E. Allen - 2006
Design of the Nulling Resistor (M8)
For the zero to be on top of the second pole (p2), the following relationship must hold
Rz = 1
gm6 CL + Cc
Cc = Cc+CL
Cc
12K’PS6I6
The resistor, Rz, is realized by the transistor M8 which is operating in the active region because the dc current through it is zero. Therefore, Rz, can be written as
Rz = vDS8iD8
|
VDS8=0=
1K’PS8(VSG8-|VTP|)
The bias circuit is designed so that voltage VA is equal to VB.
|VGS10| |VT| = |VGS8| |VT| VSG11 = VSG6 W11L11 =
I10I6
W6L6
In the saturation region
|VGS10| |VT| = 2(I10)
K'P(W10/L10) = |VGS8| |VT|
Rz = 1
K’PS8
K’PS102I10
= 1S8
S10
2K’PI10
Equating the two expressions for Rz gives W8L8 =
CcCL + Cc
S10S6I6I10
Chapter 6 – Section 3 (6/24/06) Page 6.3-15
CMOS Analog Circuit Design © P.E. Allen - 2006
Example 6.3-2 - RHP Zero Compensation
Use results of Ex. 6.3-1 and design compensation circuitry so that the RHP zero is moved from the RHP to the LHP and placed on top of the output pole p2. Use device data given in Ex. 6.3-1.
Solution
The task at hand is the design of transistors M8, M9, M10, M11, and bias current I10. The first step in this design is to establish the bias components. In order to set VA equal to VB, thenVSG11 must equal VSG6. Therefore, S11 = (I11/I6)S6
Choose I11 = I10 = I9 = 15μA which gives S11 = (15μA/95μA)94 = 14.8 15. The aspect ratio of M10 is essentially a free parameter, and will be set equal to 1. There must be sufficient supply voltage to support the sum of VSG11, VSG10, and VDS9. The ratio of I10/I5 determines the (W/L) of M9. This ratio is (W/L)9 = (I10/I5)(W/L)5 = (15/30)(4.5) = 2.25 2 Now (W/L)8 is determined to be
(W/L)8 = 3pF
3pF+10pF 1·94·95μA
15μA = 5.63 6
Chapter 6 – Section 3 (6/24/06) Page 6.3-16
CMOS Analog Circuit Design © P.E. Allen - 2006
Example 6.3-2 - Continued It is worthwhile to check that the RHP zero has been moved on top of p2. To do this, first calculate the value of Rz. VSG8 must first be determined. It is equal to VSG10, which is
VSG10 = 2I10
K’PS10 + |VTP| =
2·1550·1 + 0.7 = 1.474V
Next determine Rz.
Rz = 1
K’PS8(VSG10-|VTP|) = 106
50·5.63(1.474-.7) = 4.590k
The location of z1 is calculated as
z1 = 1
(4.590 x 103)(3x10-12) 3x10-12
942.5x10-6
= -94.46x106 rads/sec
The output pole, p2, is
p2 = -942.5x10-6
10x10-12 = -94.25x106 rads/sec
Thus, we see that for all practical purposes, the output pole is canceled by the zero that has been moved from the RHP to the LHP. The results of this design are summarized below.
W8 = 6 μm W9 = 2 μm W10 = 1 μm W11 = 15 μm
Chapter 6 – Section 3 (6/24/06) Page 6.3-17
CMOS Analog Circuit Design © P.E. Allen - 2006
An Alternate Form of Nulling Resistor
To cancel p2,
z1 = p2 Rz = Cc+CL
gm6ACC =
1gm6B
Which gives
gm6B = gm6A
Cc
Cc+CL
In the previous example, gm6A = 942.5μS, Cc = 3pF and CL = 10pF. Choose I6B = 10μA to get
gm6B = gm6ACc
Cc + CL
2KPW6BI6B
L6B =
Cc
Cc+CL
2KPW6AID6
L6A
or
W6B
L6B =
313
2 I6A
I6B W6A
L6A =
313
2 9510 (94) = 47.6 W6B = 48μm
-
+vin
M1 M2
M3 M4
M5
M6
M7
vout
VDD
VSS
VBias+
-
CcCL
M11 M10
M6B
M8 M9
Fig. 6.3-4A
Chapter 6 – Section 3 (6/24/06) Page 6.3-18
CMOS Analog Circuit Design © P.E. Allen - 2006
Programmability of the Two-Stage Op Amp The following relationships depend on the bias
current, Ibias, in the following manner and allow for programmability after fabrication.
Av(0) = gmIgmIIRIRII 1
IBias
GB = gmI
Cc IBias
Pdiss = (VDD+|VSS|)(1+K1+K2)IBias Ibias
SR = K1IBias
Cc IBias
Rout = 1
2 K2IBias
1IBias
|p1| = 1
gmIIRIRIICc
IBias2
IBias IBias
1.5
|z| = gmII
Cc IBias
Illustration of the Ibias dependence
-
+
vin
M1 M2
M3 M4
M5
M6
M7
vout
VDD
VSS
IBias
Fig. 6.3-04D
K1IBiasK2IBias
103
102
100
101
10-1
10-2
10-31 10 100
IBiasIBias(ref)
|p1|Pdiss and SR
GB and z
Ao and Rout
Fig. 160-05
Chapter 6 – Section 3 (6/24/06) Page 6.3-19
CMOS Analog Circuit Design © P.E. Allen - 2006
Simulation of the Electrical Design Area of source or drain = AS = AD = W[L1 + L2 + L3] where L1 = Minimum allowable distance between the contact in the S/D and the polysilicon (5μm) L2 = Width of a minimum size contact (5μm) L3 = Minimum allowable distance from contact in S/D to edge of S/D (5μm) AS = AD = Wx15μm Perimeter of the source or drain = PD = PS = 2W + 2(L1+L2+L3) PD = PS = 2W + 30μm Illustration:
Poly
Diffusion Diffusion
L
W
L3 L2 L1 L1 L2 L3
Fig. 6.3-5
Chapter 6 – Section 3 (6/24/06) Page 6.3-20
CMOS Analog Circuit Design © P.E. Allen - 2006
POWER SUPPLY REJECTION RATIO OF THE TWO-STAGE OP AMP
What is PSRR?
PSRR = Av(Vdd=0)Add(Vin=0)
How do you calculate PSRR? You could calculate Av and Add and divide, however
+
- VDD
VSS
Vdd
Vout
V2
V1
V2
V1
Av(V1-V2)
±AddVddVss
Vout
Fig. 180-02 Vout = AddVdd + Av(V1-V2) = AddVdd - AvVout Vout(1+Av) = AddVdd
VoutVdd
= Add
1+Av
AddAv
= 1
PSRR+ (Good for frequencies up to GB)
+
- VDD
VSS
Vdd
V2
V1Vss
VoutVin
Fig.180-01
Chapter 6 – Section 3 (6/24/06) Page 6.3-21
CMOS Analog Circuit Design © P.E. Allen - 2006
Approximate Model for PSRR+
M1 M2
M3 M4
M5
M6
M7
Vout
VDD
VSS
VBias
Cc
CII
Vdd
CI
Fig. 180-05
Cc
Vdd Rout
Vout
Other sourcesof PSRR+ besides Cc
1RoutCc
ω
VoutVdd
0dB
1.) The M7 current sink causes VSG6 to act like a battery. 2.) Therefore, Vdd couples from the source to gate of M6. 3.) The path to the output is through any capacitance from gate to drain of M6. Conclusion: The Miller capacitor Cc couples the positive power supply ripple directly to the output. Must reduce or eliminate Cc.
Chapter 6 – Section 3 (6/24/06) Page 6.3-22
CMOS Analog Circuit Design © P.E. Allen - 2006
Approximate Model for PSRR-
What is Zout?
Zout = VtIt
It = gmIIV1 = gmII
gmIVtGI+sCI+sCc
Thus, Zout = GI+s(CI+Cc)
gmIgMII
VoutVss
= 1+
rds7Zout1 =
s(Cc+CI) + GI+gmIgmIIrds7s(Cc+CI) + GI
Pole at -GI
Cc+CI
The negative PSRR is much better than the positive PSRR.
M1 M2
M3 M4
M5
M6
M7
Vout
VDD
VSS
VBias
Cc
CII
Vss
CI
Fig. 180-11
VBias connected to VSS
rds7
vout
ZoutVss
rds7
Path through Cgd7is negligible
Fig.180-12
Vtrds6||rds7
VoutCI
Cc
+
-V1
+
-gmIVoutgmIIV1RI
CII+Cgd7It
Chapter 6 – Section 4 (6/24/06) Page 6.4-1
CMOS Analog Circuit Design © P.E. Allen - 2006
SECTION 6.4 – CASCODE OP AMPS INTRODUCTION
Why Cascode Op Amps? • Control of the frequency behavior • Can get more gain by increasing the output resistance of a stage • In the past section, PSRR of the two-stage op amp was insufficient for many
applications • A two-stage op amp can become unstable for large load capacitors (if nulling resistor is
not used) • The cascode op amp leads to wider ICMR and/or smaller power supply requirements Where Should the Cascode Technique be Used? • First stage - Good noise performance Requires level translation to second stage Degrades the Miller compensation • Second stage - Self compensating Increases the efficiency of the Miller compensation Increases PSRR
Chapter 6 – Section 4 (6/24/06) Page 6.4-2
CMOS Analog Circuit Design © P.E. Allen - 2006
SINGLE STAGE CASCODE OP AMPS
Simple Single Stage Cascode Op Amp
060627-01
-
+ vin
M1 M2
M3 M4
M5
vo1
VDD
VSS
VNBias1+
-
VBias
2vin2
-
+
MC2MC1
MC4MC3
-
+ vin
M1 M2
M3 M4
M5
VDD
VSS
+
-
2vin2
-+
MC2MC1
MC4MC3
MB1 MB2
MB3 MB4
MB5
-
+
VBias
Implementation of the floating voltage VBiaswhich must equal2VON + VT.
VPBias2
VNBias1
VPBias2
Rout of the first stage is RI (gmC2rdsC2rds2)||(gmC4rdsC4rds4)
Voltage gain = vo1vin
= gm1RI [The gain is increased by approximately 0.5(gMCrdsC)]
As a single stage op amp, the compensation capacitor becomes the load capacitor.
Chapter 6 – Section 4 (6/24/06) Page 6.4-3
CMOS Analog Circuit Design © P.E. Allen - 2006
Example 6.5-1 Single-Stage, Cascode Op Amp Performance
Assume that all W/L ratios are 10 μm/1 μm, and that IDS1 = IDS2 = 50 μA of single stage op amp. Find the voltage gain of this op amp and the value of CI if GB = 10 MHz. Use the model parameters of Table 3.1-2.
Solution The device transconductances are gm1 = gm2 = gmI = 331.7 μS gmC2 = 331.7μS gmC4 = 223.6 μS. The output resistance of the NMOS and PMOS devices is 0.5 M and 0.4 M , respectively.
RI = 25 M Av(0) = 8290 V/V. For a unity-gain bandwidth of 10 MHz, the value of CI is 5.28 pF. What happens if a 100pF capacitor is attached to this op amp? GB goes from 10MHz to 0.53MHz.
Chapter 6 – Section 4 (6/24/06) Page 6.4-4
CMOS Analog Circuit Design © P.E. Allen - 2006
Enhanced Gain, Single Stage, Cascode Op Amp
060627-02
+
−vIN
vOUT
M1 M2
M3 M4
M5 M6
M7 M8
M9
VDD
VNB1
-A
-A-A
+
−vIN
vOUT
M1 M2
M3 M4
M5 M6
M7 M8
M9
VDD
VNB1 M10
VNB1
VDD VDDVPB1
M11 M12
M13 M14
M15
M16
From inspection, we can write the voltage gain as,
Av = vOUTvIN = gm1Rout where Rout = (Ards6gm6rds8)|| (Ards2gm4rds4)
Since A gmrds the voltage gain would be equal to 100,000 to 500,000.
Chapter 6 – Section 4 (6/24/06) Page 6.4-5
CMOS Analog Circuit Design © P.E. Allen - 2006
TWO-STAGE, CASCODE OP AMPS Two-Stage Op Amp with a Cascoded First-Stage
070117-01
Cc
-
+vin
M1 M2
M3 M4
M5
vo1
VDD
VSS
VBias+
-
R
2vin2-
+
MC2MC1
MC4MC3
MB1 MB2
MB3 MB4
MB5
-
+
VBias
MT1
MT2
M6
M7
vout
• MT1 and MT2 are required for level shifting from
the first-stage to the second. • The PSRR+ is improved by the presence of MT1 • Internal loop pole at the gate of M6 may cause the
Miller compensation to fail. • The voltage gain of this op amp could easily be 100,000V/V
σ
jω
z1p1p2p3Fig. 6.5-2A
Chapter 6 – Section 4 (6/24/06) Page 6.4-6
CMOS Analog Circuit Design © P.E. Allen - 2006
Two-Stage Op Amp with a Cascode Second-Stage
Av = gmIgmIIRIRII where gmI = gm1 = gm2, gmII = gm6,
RI = 1
gds2 + gds4 = 2
( 2 + 4)ID5
and RII = (gmC6rdsC6rds6)||(gmC7rdsC7rds7) Comments: • The second-stage gain has greatly increased improving the Miller compensation • The overall gain is approximately (gmrds)3 or very large • Output pole, p2, is approximately the same if Cc is constant • The zero RHP is the same if Cc is constant • PSRR is poor unless the Miller compensation is removed (then the op amp becomes
self compensated)
-
+
vin
M1 M2
M3 M4
M5
M6
M7
vout
VDD
VSS
VBias+
-
Cc
CL
VBP
VBN
MC6
MC7
Fig. 6.5-3
Rz
Chapter 6 – Section 4 (6/24/06) Page 6.4-7
CMOS Analog Circuit Design © P.E. Allen - 2006
A Balanced, Two-Stage Op Amp using a Cascode Output Stage
vout = gm1gm8
gm3 vin2 +
gm2gm6 gm4
vin2 RII
= gm1
2 +gm2
2 kvin RII = gm1·k·RII vin
where RII = (gm7rds7rds6)||(gm12rds12rds11) and
k = gm8gm3 =
gm6gm4
This op amp is balanced because the drain-to-ground loads for M1 and M2 are identical. TABLE 1 - Design Relationships for Balanced, Cascode Output Stage Op Amp.
Slew rate = Iout
CL GB =
gm1gm8
gm3CL Av =
12
gm1gm8
gm3 +
gm2gm6
gm4 RII
Vin(max) = VDD I5
3
1/2 |VTO3|(max) +VT1(min) Vin(min) = VSS + VDS5 +
I5
1
1/2 + VT1(min)
060627-03
-
+
vinM1 M2
M3
M4
M5
M6
M11
vout
VDD
VSS
+
-
CLM9
M10
M8
M12
M7VPB2
VNB2
VNB1
Chapter 6 – Section 4 (6/24/06) Page 6.4-8
CMOS Analog Circuit Design © P.E. Allen - 2006
Example 6.5-2 Design of Balanced, Cascoded Output Stage Op Amp
The balanced, cascoded output stage op amp is a useful alternative to the two-stage op amp. Its design will be illustrated by this example. The pertinent design equations for the op amp were given above. The specifications of the design are as follows: VDD = VSS = 2.5 V Slew rate = 5 V/μs with a 50 pF load GB = 10 MHz with a 25 pF load Av 5000 Input CMR = 1V to +1.5 V Output swing = ±1.5 V Use the parameters of Table 3.1-2 and let all device lengths be 1 μm.
Solution While numerous approaches can be taken, we shall follow one based on the above specifications. The steps will be numbered to help illustrate the procedure. 1.) The first step will be to find the maximum source/sink current. This is found from the slew rate. Isource/Isink = CL slew rate = 50 pF(5 V/μs) = 250 μA 2.) Next some W/L constraints based on the maximum output source/sink current are developed. Under dynamic conditions, all of I5 will flow in M4; thus we can write Max. Iout(source) = (S6/S4)I5 and Max. Iout(sink) = (S8/S3)I5 The maximum output sinking current is equal to the maximum output sourcing current if
S3 = S4, S6 = S8, and S10 = S11
Chapter 6 – Section 4 (6/24/06) Page 6.4-9
CMOS Analog Circuit Design © P.E. Allen - 2006
Example 6.5-2 - Continued
3.) Choose I5 as 100 μA. This current (which can be changed later) gives S6 = 2.5S4 and S8 = 2.5S3 Note that S8 could equal S3 if S11 = 2.5S10. This would minimize the power dissipation.
4.) Next design for ±1.5 V output capability. We shall assume that the output must source or sink the 250μA at the peak values of output. First consider the negative output peak. Since there is 1 V difference between VSS and the minimum output, let VDS11(sat) = VDS12(sat) = 0.5 V (we continue to ignore the bulk effects). Under the maximum negative peak assume that I11 = I12 = 250 μA. Therefore
0.5 = 2I11
K'NS11 = 2I12
K'NS12 = 500 μA
(110 μA/V2)S11
which gives S11 = S12 = 18.2 and S9 = S10 = 18.2. For the positive peak, we get
0.5 = 2I6
K'PS6 = 2I7
K'PS7 = 500 μA
(50 μA/V2)S6
which gives S6 = S7 = S8 = 40 and S3 = S4 = (40/2.5) = 16.
Chapter 6 – Section 4 (6/24/06) Page 6.4-10
CMOS Analog Circuit Design © P.E. Allen - 2006
Example 6.5-2 - Continued 5.) Now we must consider the possibility of conflict among the specifications. First consider the input CMR. S3 has already been designed as 16. Using ICMR relationship, we find that S3 should be at least 4.1. A larger value of S3 will give a higher value of Vin(max) so that we continue to use S3 = 16 which gives Vin(max) = 1.95V. Next, check to see if the larger W/L causes a pole below the gainbandwidth. Assuming a Cox of 0.4fF/μm2 gives the first-stage pole of
p3 = -gm3
Cgs3+Cgs8 =
- 2K’PS3I3(0.667)(W3L3+W8L8)Cox
= 33.15x109 rads/sec or 5.275GHz
which is much greater than 10GB. 6.) Next we find gm1 (gm2). There are two ways of calculating gm1. (a.) The first is from the Av specification. The gain is Av = (gm1/2gm4)(gm6 + gm8) RII Note, a current gain of k could be introduced by making S6/S4 (S8/S3 = S11/S3) equal to k.
gm6gm4
= gm11gm3
= 2KP’·S6·I62KP’·S4·I4 = k
Calculating the various transconductances we get gm4 = 282.4 μS, gm6 = gm7 = gm8 = 707 μS, gm11 = gm12 = 707 μS, rds6 = rd7 = 0.16 M , and rds11 = rds12 = 0.2 M . Assuming that the gain Av must be greater than 5000 and k = 2.5 gives gm1 > 72.43 μS.
Chapter 6 – Section 4 (6/24/06) Page 6.4-11
CMOS Analog Circuit Design © P.E. Allen - 2006
Example 6.5-2 - Continued
(b.) The second method of finding gm1 is from the GB specifications. Multiplying the gain by the dominant pole (1/CIIRII) gives
GB = gm1(gm6 + gm8)
2gm4CL
Assuming that CL= 25 pF and using the specified GB gives gm1 = 251 μS. Since this is greater than 72.43μS, we choose gm1 = gm2 = 251μS. Knowing I5 gives S1 = S2 = 5.7 6. 8.) The next step is to check that S1 and S2 are large enough to meet the 1V input CMR specification. Use the saturation formula we find that VDS5 is 0.261 V. This gives S5 = 26.7 27. The gain becomes Av = 6,925V/V and GB = 10 MHz for a 25 pF load. We shall assume that exceeding the specifications in this area is not detrimental to the performance of the op amp. 9.) Knowing the currents and W/L values, the bias voltages VNB1, VNB2 and VPB2 can be designed. The W/L values resulting from this design procedure are shown below. The power dissipation for this design is seen to be 2 mW. The next step would be simulation. S1 = S2 = 6 S3 = S4 = 16 S5 = 27 S6 = S7 = S8 = 40 S9 = S10 = S11 = S12 = 18.2
Chapter 6 – Section 4 (6/24/06) Page 6.4-12
CMOS Analog Circuit Design © P.E. Allen - 2006
Technological Implications of the Cascode Configuration
Fig. 6.5-5
�� ��������Poly IIPoly I
n+ n+n-channel
p substrate/well
A B C D
Thinoxide
A
B
C
D
If a double poly CMOS process is available, internode parasitics can be minimized. As an alternative, one should keep the drain/source between the transistors to a minimum area.
Fig. 6.5-5A
�� �����Poly I
n+ n+n-channel
p substrate/well
A B C D
Thinoxide
A
B
C
D
Poly I
Minimum Polyseparation
����n-channeln+
Chapter 6 – Section 4 (6/24/06) Page 6.4-13
CMOS Analog Circuit Design © P.E. Allen - 2006
Input Common Mode Range for Two Types of Differential Amplifier Loads
vicm
M1 M2
M3 M4
M5
VDD
VSS
VBias+
-
+
-
VSG3
M1 M2
M3 M4
M5
VDD
VSS
VBias+
-
+
-
VSD3
VBP
+
-
VSD4
+
-
VSD4
VDD-VSG3+VTN
VSS+VDS5+VGS1
InputCommon
ModeRange
vicm
VDD-VSD3+VTN
VSS+VDS5+VGS1
InputCommon
ModeRange
Differential amplifier witha current mirror load. Fig. 6.5-6
Differential amplifier withcurrent source loads.
In order to improve the ICMR, it is desirable to use current source (sink) loads without losing half the gain. The resulting solution is the folded cascode op amp.
Chapter 6 – Section 4 (6/24/06) Page 6.4-14
CMOS Analog Circuit Design © P.E. Allen - 2006
The Folded Cascode Op Amp
Comments: • I4 and I5, should be designed so that I6 and I7 never become zero (i.e. I4=I5=1.5I3) • This amplifier is nearly balanced (would be exactly if RA was equal to RB) • Self compensating • Poor noise performance, the gain occurs at the output so all intermediate transistors
contribute to the noise along with the input transistors. (Some first stage gain can be achieved if RA and RB are greater than gm1 or gm2.
060628-04
VPB1
M4 M5
RAI6
VPB2 RB
I4 I5
VDD
I7M6 M7
VNB2
M8 M9
M10 M11
+−vIN
vOUT
VNB1
I1 I2
M1 M2
M3I3
CL
Chapter 6 – Section 4 (6/24/06) Page 6.4-15
CMOS Analog Circuit Design © P.E. Allen - 2006
Small-Signal Analysis of the Folded Cascode Op Amp Model:
Recalling what we learned about the resistance looking into the source of the cascode transistor;
RA = rds6+(1/gm10)1 + gm6rds6
1
gm6 and RB =
rds7 + RII1 + gm7rds7
RII
gm7rds7 where RII gm9rds9rds11
The small-signal voltage transfer function can be found as follows. The current i10 is written as
i10 = -gm1(rds1||rds4)vin2[RA + (rds1||rds4)]
-gm1vin2
and the current i7 can be expressed as
i7 = gm2(rds2||rds5)vin
2RII
gm7rds7 + (rds2||rds5)
= gm2vin
2 1 + RII(gds2+gds5)
gm7rds7
= gm2vin2(1+k) where k =
RII(gds2+gds5)gm7rds7
060628-02
gm1vin2 rds1 rds4
rds6
gm6vgs6RA
gm2vin2 rds2 rds5
rds7
gm7vgs7
RII
RB
i10
i10+
-vgs7
+
-vgs6
+
-vout
i7
1gm10
The output voltage, vout, is equal to the sum of i7 and i10 flowing through Rout. Thus,
voutvin
= gm12 +
gm22(1+k) Rout =
2+k2+2k gmIRout
Chapter 6 – Section 4 (6/24/06) Page 6.4-16
CMOS Analog Circuit Design © P.E. Allen - 2006
Frequency Response of the Folded Cascode Op Amp
The frequency response of the folded cascode op amp is determined primarily by the output pole which is given as
pout = -1
RoutCout
where Cout is all the capacitance connected from the output of the op amp to ground. All other poles must be greater than GB = gm1/Cout. The approximate expressions for each pole is 1.) Pole at node A: pA - gm6/CA 2.) Pole at node B: pB - gm7/CB 3.) Pole at drain of M6: p6 -gm10/C6 4.) Pole at source of M8: p8 -(gm8rds8gm10)/C8 5.) Pole at source of M9: p9 -gm9/C9
where the approximate expressions are found by the reciprocal product of the resistance and parasitic capacitance seen to ground from a given node. One might feel that because RB is approximately rds that this pole might be too small. However, at frequencies where this pole has influence, Cout, causes Rout to be much smaller making pB also non-dominant.
Chapter 6 – Section 4 (6/24/06) Page 6.4-17
CMOS Analog Circuit Design © P.E. Allen - 2006
Example 6.5-3 - Folded Cascode, CMOS Op Amp
Assume that all gmN = gmP = 100μS, rdsN = 2M , rdsP = 1M , and CL = 10pF. Find all of the small-signal performance values for the folded-cascode op amp.
RII = 0.4G , RA = 10k , and RB = 4M k = 0.4x109(0.3x10-6)
100 = 1.2
voutvin
= 2+1.22+2.4 (100)(57.143) = 4,156V/V
Rout = RII ||[gm7rds7(rds5||rds2)] = 400M ||[(100)(0.667M )] = 57.143M
|pout| = 1
RoutCout =
157.143M ·10pF = 1,750 rads/sec. 278Hz GB = 1.21MHz
Chapter 6 – Section 4 (6/24/06) Page 6.4-18
CMOS Analog Circuit Design © P.E. Allen - 2006
PSRR of the Folded Cascode Op Amp Consider the following circuit used to model the PSRR-:
Vout
Vss
Cgd11
M11
M9
VDD
R
Fig. 6.5-9A
Vss Vout
Cgd9
Rout
+
-
Cgd9
VGS11
VGSG9
Vss
Vss
Vss
rds11
Cout
rds9
This model assumes that gate, source and drain of M11 and the gate and source of M9 all vary with VSS.
We shall examine Vout/Vss rather than PSRR-. (Small Vout/Vss will lead to large PSRR-.) The transfer function of Vout/Vss can be found as
VoutVss
sCgd9Rout
sCoutRout+1 for Cgd9 < Cout
The approximate PSRR- is sketched on the next page.
Chapter 6 – Section 4 (6/24/06) Page 6.4-19
CMOS Analog Circuit Design © P.E. Allen - 2006
Frequency Response of the PSRR- of the Folded Cascode Op Amp
VoutVss
GB
Cgd9Rout
|PSRR-|
dB
0dB
Fig. 6.5-10A
log10(ω)
1
Cout
Cgd9
Other sources of Vss injection, i.e. rds9
Dominantpole frequency
|Avd(ω)|
We see that the PSRR of the cascode op amp is much better than the two-stage op amp.
Chapter 6 – Section 4 (6/24/06) Page 6.4-20
CMOS Analog Circuit Design © P.E. Allen - 2006
Design Approach for the Folded-Cascode Op Amp Step Relationship Design Equation/Constraint Comments
1 Slew Rate I3 = SR·CL
2 Bias currents in output cascodes
I4 = I5 = 1.2I3 to 1.5I3 Avoid zero current in cascodes
3 Maximum output voltage, vout(max)
S5=2I5
KP’VSD52 , S7=2I7
KP’VSD72 , (S4=S5 and S6= S7) VSD5(sat)=VSD7(sat)
= 0.5[VDD-Vout(max)]
4 Minimum output voltage, vout(min)
S11=2I11
KN’VDS112 , S9=2I9
KN’VDS92 , (S10=S11and S8=S9) VDS9(sat)=VDS11(sat)
= 0.5[Vout(min)-VSS]
5 GB =
gm1CL S1=S2=
gm12
KN’I3 = GB2CL2
KN’I3
6 Minimum input CM S3 =
2I3
KN’ Vin(min)-VSS- (I3/KN’S1) -VT1 2
7 Maximum input CM S4 = S5 =
2I4KP’( )VDD-Vin(max)+VT1
2 S4 and S5 must meet or
exceed value in step 3
8 Differential Voltage Gain
voutvin =
gm12 +
gm22(1+k) Rout =
2+k2+2k gmIRout k =
RII(gds2+gds4)gm7rds7
9 Power dissipation Pdiss = (VDD-VSS)(I3+I10+I11)
Chapter 6 – Section 4 (6/24/06) Page 6.4-21
CMOS Analog Circuit Design © P.E. Allen - 2006
Example 6.5-3 Design of a Folded-Cascode Op Amp Follow the procedure given to design the folded-cascode op amp when the slew rate is 10V/μs, the load capacitor is 10pF, the maximum and minimum output voltages are ±2V for ±2.5V power supplies, the GB is 10MHz, the minimum input common mode voltage is -1.5V and the maximum input common mode voltage is 2.5V. The differential voltage gain should be greater than 5,000V/V and the power dissipation should be less than 5mW. Use channel lengths of 1μm. Solution Following the approach outlined above we obtain the following results. I3 = SR·CL = 10x106·10-11 = 100μA Select I4 = I5 = 125μA. Next, we see that the value of 0.5(VDD-Vout(max)) is 0.5V/2 or 0.25V. Thus,
S4 = S5 = 2·125μA
50μA/V2·(0.25V)2 = 2·125·16
50 = 80
and assuming worst case currents in M6 and M7 gives,
S6 = S7 = 2·125μA
50μA/V2(0.25V)2 = 2·125·16
50 = 80
The value of 0.5(Vout(min)-|VSS|) is also 0.25V which gives the value of S8, S9, S10 and
S11 as S8 = S9 = S10 = S11 = 2·I8
KN’VDS82 = 2·125
110·(0.25)2 = 36.36
Chapter 6 – Section 4 (6/24/06) Page 6.4-22
CMOS Analog Circuit Design © P.E. Allen - 2006
Example 6.5-3 - Continued
In step 5, the value of GB gives S1 and S2 as
S1 = S2 = GB2·CL2
KN’I3 =
(20 x106)2(10-11)2
110x10-6·100x10-6 = 35.9
The minimum input common mode voltage defines S3 as
S3 = 2I3
KN’ Vin(min)-VSS-I3
KN’S1 - VT1
2 =
200x10-6
110x10-6 -1.5+2.5-100
110·35.9 -0.7 2 = 91.6
We need to check that the values of S4 and S5 are large enough to satisfy the maximum input common mode voltage. The maximum input common mode voltage of 2.5 requires
S4 = S5 2I4
KP’[VDD-Vin(max)+VT1]2 = 2·125μA
50x10-6μA/V2[0.7V]2 = 10.2
which is much less than 80. In fact, with S4 = S5 = 80, the maximum input common mode voltage is 3V.
The power dissipation is found to be
Pdiss = 5V(125μA+125μA) = 1.25mW
Chapter 6 – Section 4 (6/24/06) Page 6.4-23
CMOS Analog Circuit Design © P.E. Allen - 2006
Example 6.5-3 - Continued
The small-signal voltage gain requires the following values to evaluate: S4, S5: gm = 2·125·50·80 = 1000μS and gds = 125x10-6·0.05 = 6.25μS
S6, S7: gm = 2·75·50·80 = 774.6μS and gds = 75x10-6·0.05 = 3.75μS
S8, S9, S10, S11: gm = 2·75·110·36.36 = 774.6μS and gds = 75x10-6·0.04 = 3μS
S1, S2: gmI = 2·50·110·35.9 = 628μS and gds = 50x10-6(0.04) = 2μS
Thus,
RII gm9rds9rds11 = (774.6μS)1
3μS1
3μS = 86.07M
Rout 86.07M ||(774.6μS)1
3.75μS1
2μS+6.25μS = 19.40M
k = RII(gds2+gds4)
gm7rds7 =
86.07M (2μS+6.25μS)(3.75μS)774.6μS = 3.4375
The small-signal, differential-input, voltage gain is
Avd = 2+k2+2k gmIRout =
2+3.43752+6.875 0.628x10-3·19.40x106 = 7,464 V/V
The gain is larger than required by the specifications which should be okay.
Chapter 6 – Section 4 (6/24/06) Page 6.4-24
CMOS Analog Circuit Design © P.E. Allen - 2006
Comments on Folded Cascode Op Amps
• Good PSRR
• Good ICMR
• Self compensated
• Can cascade an output stage to get extremely high gain with lower output resistance (use Miller compensation in this case)
• Need first stage gain for good noise performance
• Widely used in telecommunication circuits where large dynamic range is required
Chapter 6 – Section 4 (6/24/06) Page 6.4-25
CMOS Analog Circuit Design © P.E. Allen - 2006
Enhanced-Gain, Folded Cascode Op Amps
If more gain is needed, the folded cascode op amp can be enhanced to boost the output impedance even higher as follows.
Voltage gain = gm1Rout,
where Rout [Ards7gm7(rds1||rds5)]|| (Ards9gm9rds11)
Since A gmrds the voltage gain would be in the range of 100,000 to 500,000.
Note that to achieve maximum output swing, it will be necessary to make sure that M5 and M11 are biased with VDS = VDS(sat).
060718-03
vOUT
M4M5
M3
M7
M8 M9
M10 M11
M6
VDD
VPB1
-A
-A-A
+
−vIN
M1 M2
VNB1
Chapter 6 – Section 4 (6/24/06) Page 6.4-26
CMOS Analog Circuit Design © P.E. Allen - 2006
What are the Enhancement Amplifiers? Requirements: 1.) Need a gain of gmrds.
2.) Must be able to set the dc voltage at its input to get wide-output voltage swing.
Possible Enhancement Amplifiers:
060718-05
VDDVPB1
VPB2
VNB1VDD-VSD(Sat)
vin
vout
M1 M2
M3
M4
M5
M6
-A
vin
vout
VDDVPB1
VNB2
VNB1
vin
vout
VDS(Sat)M1 M2
M3
M4
M5
M6
vout
-A
vin
Chapter 6 – Section 4 (6/24/06) Page 6.4-27
CMOS Analog Circuit Design © P.E. Allen - 2006
Frequency Response of the Enhanced Gain Cascode Op Amps
Normally, the frequency response of the cascode op amps would have one dominant pole at the output. The frequency response would be,
Av(s) = gm1Rout(1/sCout) Rout +1/sCout =
gm1Rout sRoutCout +1 =
gm1Rout
1 - s
p1
If the amplifier used to boost the output resistance had no frequency dependence then the frequency response would be as follows.
060629-02
0dB
40dB
60dB
80dB
100dBEnhanced Gain Cascode Op Amp
Normal CascodeOp Amp
|p1(enh)| GBlog10ω
Gain (db)
|p1|
Chapter 6 – Section 4 (6/24/06) Page 6.4-28
CMOS Analog Circuit Design © P.E. Allen - 2006
Frequency Response of the Enhanced Gain Cascode Op Amp – Continued
• Does the pole in the feedback amplifier A have an influence?
Although the output resistance can be modeled as,
Rout’ RoutAo
1- s
p2Ao
1- s
p2
it has no influence on the frequency response because Cout has shorted out any influence a change in Rout might have.
• Higher order poles come from a diversion of the current flow in the op amp to ground rather than the intended destination of the current to the output. These poles that divert the current are:
- Pole at the source of M6 (Agm6/C6) - Pole at the source of M7 (Agm7/C7) - Pole at the drain of M8 (Agm10/C8) - Pole at the source of M9 (Agm9/C6) - Pole at the drain of M10 (gm8rds8gm10/C10)
Note that the enhancement amplifiers cause all higher-order poles to be moved out by |A|. Therefore, the GB of the enhanced gain cascode op amp should be able to be increased by |A|.
Chapter 6 – Section 5 (6/24/06) Page 6.5-1
CMOS Analog Circuit Design © P.E. Allen - 2006
SECTION 6.5 - SIMULATION AND MEASUREMENT OF OP AMPS Simulation and Measurement Considerations
Objectives:
• The objective of simulation is to verify and optimize the design. • The objective of measurement is to experimentally confirm the specifications.
Similarity Between Simulation and Measurement:
• Same goals • Same approach or technique
Differences Between Simulation and Measurement:
• Simulation can idealize a circuit • Measurement must consider all nonidealities
Chapter 6 – Section 5 (6/24/06) Page 6.5-2
CMOS Analog Circuit Design © P.E. Allen - 2006
Simulating or Measuring the Open-Loop Transfer Function of the Op Amp
Circuit (Darkened op amp identifies the op amp under test):
Simulation:
This circuit will give the voltage transfer function curve. This curve should identify:
1.) The linear range of operation 2.) The gain in the linear range 3.) The output limits 4.) The systematic input offset voltage 5.) DC operating conditions, power dissipation 6.) When biased in the linear range, the small-signal frequency response can be
obtained 7.) From the open-loop frequency response, the phase margin can be obtained (F = 1)
Measurement:
This circuit probably will not work unless the op amp gain is very low.
Fig. 240-01
+ -VOSvIN
vOUTVDD
VSSRLCL
Chapter 6 – Section 5 (6/24/06) Page 6.5-3
CMOS Analog Circuit Design © P.E. Allen - 2006
A More Robust Method of Measuring the Open-Loop Frequency Response
Circuit:
vIN vOUT
VDD
VSSRLCLRC
Fig. 240-02 Resulting Closed-Loop Frequency Response:
dB
log10(w)
Av(0)
1RC RC
Av(0)
Op AmpOpen LoopFrequencyResponse
Fig. 240-03
0dB
Make the RC product as large as possible.
Chapter 6 – Section 5 (6/24/06) Page 6.5-4
CMOS Analog Circuit Design © P.E. Allen - 2006
Simulation and Measurement of Open-Loop Frequency Response with Moderate Gain Op Amps
vIN vOUTVDD
VSSRLCL
R
R
+
-
vi
Fig. 240-04
Make R as large and measure vout and vi to get the open loop gain.
Chapter 6 – Section 5 (6/24/06) Page 6.5-5
CMOS Analog Circuit Design © P.E. Allen - 2006
Simulation of the Common-Mode Voltage Gain
VOSvout
VDD
VSSRLCL
+ -
vcm
+
-
Fig. 6.6-5
Make sure that the output voltage of the op amp is in the linear region.
Chapter 6 – Section 5 (6/24/06) Page 6.5-6
CMOS Analog Circuit Design © P.E. Allen - 2006
Simulation of CMRR of an Op Amp None of the above methods are really suitable for simulation of CMRR. Consider the following:
VDD
VSS
Vcm
Vout
V2
V1
V2
V1
Av(V1-V2)
±AcVcmVcm
Vout
Vcm
Vcm
+
-
Fig. 6.6-7
Vout = Av(V1-V2) ±Acm
V1+V2
2 = -AvVout ± AcmVcm
Vout = ±Acm
1+Av Vcm
±Acm
Av Vcm
|CMRR| = Av
Acm =
Vcm
Vout
(However, PSRR+ must equal PSRR-)
Chapter 6 – Section 5 (6/24/06) Page 6.5-7
CMOS Analog Circuit Design © P.E. Allen - 2006
Direct Simulation of PSRR
Circuit:
VDD
VSS
Vdd
Vout
V2
V1
V2
V1
Av(V1-V2)
±AddVddVss
Vss = 0
Fig. 6.6-9
+
-
Vout = Av(V1-V2) ±AddVdd = -AvVout ± AddVdd
Vout = ±Add
1+Av Vdd
±Add
Av Vdd
PSRR+ = Av
Add =
Vdd
Vout and PSRR- =
Av
Ass =
Vss
Vout
Works well as long as CMRR is much greater than 1.
Chapter 6 – Section 5 (6/24/06) Page 6.5-8
CMOS Analog Circuit Design © P.E. Allen - 2006
Measurement of Av, CMRR and PSRR Configuration:
vOS
vOUTVDD
VSSRLCL
+
-
+- 100kΩ
100kΩ
10kΩ
10Ω
vSET
vI
Fig. 240-07 Note: 1.) PSRR- can be measured similar to
PSRR+ by changing only VSS. 2.) The ±1V perturbation can be
replaced by a sinusoid to measure CMRR or PSRR as follows:
PSRR+ = 1000·vdd
vos , PSRR- =
1000·vssvos
and CMRR = 1000·vcm
vos
Note that vI vOS1000 or vOS 1000vI
How Does this Circuit Work?
CMRR: PSRR: 1.) Set VDD’ = VDD + 1V VSS’ = VSS + 1V vOUT’ = vOUT + 1V 2.) Measure vOS (vOS1)
3.) Set VDD’ = VDD - 1V VSS’ = VSS - 1V vOUT’ = vOUT - 1V 4.) Measure vOS (vOS2) 5.)
CMRR=2000
|vOS2-vOS1|
1.) Set VDD’ = VDD + 1V VSS’ = VSS vOUT’ = 0V 2.) Measure vOS (vOS3) 3.) Set VDD’ = VDD - 1V VSS’ = VSS vOUT’ = 0V 4.) Measure vOS (vOS4) 5.)
PSRR+=2000
|vOS4-vOS3|
Chapter 6 – Section 5 (6/24/06) Page 6.5-9
CMOS Analog Circuit Design © P.E. Allen - 2006
How Does the Previous Idea Work? A circuit is shown which is used to measure the CMRR and PSRR of an op amp. Prove that the CMRR can be given as
CMRR = 1000 vicm
vos
Solution The definition of the common-mode rejection ratio is
CMRR = Avd
Acm =
(vout/vid)(vout/vicm)
However, in the above circuit the value of vout is the same so that we get
CMRR = vicm
vid
But vid = vi and vos 1000vi = 1000vid vid = vos
1000
Substituting in the previous expression gives, CMRR = vicm
vos
1000
= 1000 vicm
vos
vos
vOUTVDD
VSSRLCL
+
-
+- 100kΩ
100kΩ
10kΩ
10Ω
vicm
vi
Fig. 240-08
vicm
Chapter 6 – Section 5 (6/24/06) Page 6.5-10
CMOS Analog Circuit Design © P.E. Allen - 2006
Measurement of Open-Loop Gain
Measurement configuration:
060701-01
Vos
VDD
VSSRLCL
+
-
+- 100kΩ
100kΩ
10kΩ
10Ω
Vout
Vi
Vout
Avd = VoutVid =
VoutVi
Vos 1000Vi
Therefore, Avd = 1000Vout
Vos
Chapter 6 – Section 5 (6/24/06) Page 6.5-11
CMOS Analog Circuit Design © P.E. Allen - 2006
Simulation or Measurement of ICMR
vIN
vOUTVDD
VSSRLCL
+
-
Fig.240-11
ICMR
IDD
vOUT
vIN
1
1
Also, monitor IDD or ISS.
ISS
Initial jump in sweep is due to the turn-on of M5.
Should also plot the current in the input stage (or the power supply current).
Chapter 6 – Section 5 (6/24/06) Page 6.5-12
CMOS Analog Circuit Design © P.E. Allen - 2006
Measurement or Simulation of Slew Rate and Settling Time
vin
voutVDD
VSSRLCL
+
-
IDD Settling ErrorTolerance
1
+SR
1
-SR
Peak Overshoot
Feedthrough
vin
vout
Settling Time
Volts
t
Fig. 240-14 If the slew rate influences the small signal response, then make the input step size small enough to avoid slew rate (i.e. less than 0.5V for MOS).
Chapter 6 – Section 5 (6/24/06) Page 6.5-13
CMOS Analog Circuit Design © P.E. Allen - 2006
Phase Margin and Peak Overshoot Relationship It can be shown (Appendix C of the text) that:
Phase Margin (Degrees) = 57.2958cos-1[ 4 4+1 - 2 2] Overshoot (%)
= 100 exp-
1- 2
For example, a 5% overshoot corresponds to a phase margin of approximately 64°.
0
10
20
30
40
50
60
70
80
Phas
e M
argi
n (D
egre
es)
1.0
10
0 0.2 0.4 0.6 0.8 1
Ove
rsho
ot (
%)
Phase Margin Overshoot
100
0.1
ζ= 12Q
Fig. 240-15
Chapter 6 – Section 5 (6/24/06) Page 6.5-14
CMOS Analog Circuit Design © P.E. Allen - 2006
Example 6.5-1 Simulation of the CMOS Op Amp of Ex. 6.3-1. The op amp designed in Example 6.3-1 and shown in Fig. 6.3-3 is to be analyzed by SPICE to determine if the specifications are met. The device parameters to be used are those of Tables 3.1-2 and 3.2-1. In addition to verifying the specifications of Example 6.3-1, we will simulate PSRR+ and PSRR-. Solution/Simulation The op amp will be treated as a subcircuit in order to simplify the repeated analyses. The table on the next page gives the SPICE subcircuit description of Fig. 6.3-3. While the values of AD, AS, PD, and PS could be calculated if the physical layout was complete, we will make an educated estimate of these values by using the following approximations. AS = AD W[L1 + L2 + L3] PS = PD 2W + 2[L1 + L2 + L3]
-
+
vin
M1 M2
M3 M4
M5
M6
M7
vout
VDD = 2.5V
VSS = -2.5V
Cc = 3pF
CL =10pF
3μm1μm
3μm1μm
15μm1μm
15μm1μm
M84.5μm1μm
30μA
4.5μm1μm
14μm1μm
94μm1μm
30μA
95μA
Fig. 240-16
where L1 is the minimum allowable distance between the polysilicon and a contact in the moat (Rule 5C of Table 2.6-1), L2 is the length of a minimum-size square contact to moat (Rule 5A of Table 2.6-1), and L3 is the minimum allowable distance between a contact to moat and the edge of the moat (Rule 5D of Table 2.6-1).
Chapter 6 – Section 5 (6/24/06) Page 6.5-15
CMOS Analog Circuit Design © P.E. Allen - 2006
Example 6.6-2 - Continued Op Amp Subcircuit:
-
+vin vout
VDD
VSS
+
-6
2
1
8
9 Fig. 240-17
.SUBCKT OPAMP 1 2 6 8 9
M1 4 2 3 3 NMOS1 W=3U L=1U AD=18P AS=18P PD=18U PS=18U
M2 5 1 3 3 NMOS1 W=3U L=1U AD=18P AS=18P PD=18U PS=18U
M3 4 4 8 8 PMOS1 W=15U L=1U AD=90P AS=90P PD=42U PS=42U
M4 5 4 8 8 PMOS1 W=15U L=1U AD=90P AS=90P PD=42U PS=42U
M5 3 7 9 9 NMOS1 W=4.5U L=1U AD=27P AS=27P PD=21U PS=21U
M6 6 5 8 8 PMOS1 W=94U L=1U AD=564P AS=564P PD=200U PS=200U
M7 6 7 9 9 NMOS1 W=14U L=1U AD=84P AS=84P PD=40U PS=40U
M8 7 7 9 9 NMOS1 W=4.5U L=1U AD=27P AS=27P PD=21U PS=21U
CC 5 6 3.0P
.MODEL NMOS1 NMOS VTO=0.70 KP=110U GAMMA=0.4 LAMBDA=0.04 PHI=0.7
+MJ=0.5 MJSW=0.38 CGBO=700P CGSO=220P CGDO=220P CJ=770U CJSW=380P
+LD=0.016U TOX=14N
.MODEL PMOS1 PMOS VTO=-0.7 KP=50U GAMMA=0..57 LAMBDA=0.05 PHI=0.8
+MJ=0.5 MJSW=.35 CGBO=700P CGSO=220P CGDO=220P CJ=560U CJSW=350P +LD=0.014U TOX=14N
IBIAS 8 7 30U
.ENDS
Chapter 6 – Section 5 (6/24/06) Page 6.5-16
CMOS Analog Circuit Design © P.E. Allen - 2006
Example 6.6-2 - Continued PSPICE Input File for the Open-Loop Configuration: EXAMPLE 1 OPEN LOOP CONFIGURATION
.OPTION LIMPTS=1000
VIN+ 1 0 DC 0 AC 1.0
VDD 4 0 DC 2.5
VSS 0 5 DC 2.5
VIN - 2 0 DC 0
CL 3 0 10P
X1 1 2 3 4 5 OPAMP . . . (Subcircuit of previous slide) . . . .OP
.TF V(3) VIN+
.DC VIN+ -0.005 0.005 100U
.PRINT DC V(3)
.AC DEC 10 1 10MEG
.PRINT AC VDB(3) VP(3)
.PROBE (This entry is unique to PSPICE)
.END
Chapter 6 – Section 5 (6/24/06) Page 6.5-17
CMOS Analog Circuit Design © P.E. Allen - 2006
Example 6.6-2 - Continued Open-loop transfer characteristic of Example 6.3-1:
-2
-1
0
1
2
-2 -1.5 -1.0 -0.5 0 0.5 1 1.5 2
v OU
T(V
)
vIN(mV)
2.5
-2.5
VOS
Fig. 240-18
Chapter 6 – Section 5 (6/24/06) Page 6.5-18
CMOS Analog Circuit Design © P.E. Allen - 2006
Example 6.6-2 - Continued Open-loop transfer frequency response of Example 6.3-1:
-40
-20
0
20
40
60
80
10 100 1000 104 105 106 107 108
Mag
nitu
de (
dB)
Frequency (Hz)
GB-200
-150
-100
-50
0
50
100
150
200
10 100 1000 104 105 106 107 108
Phas
e Sh
ift (
Deg
rees
)
Frequency (Hz)
GBPhase Margin
Fig. 6.6-16
Chapter 6 – Section 5 (6/24/06) Page 6.5-19
CMOS Analog Circuit Design © P.E. Allen - 2006
Example 6.6-2 - Continued Input common mode range of Example 6.3-1: EXAMPLE 6.6-1 UNITY GAIN CONFIGURATION.
.OPTION LIMPTS=501
VIN+ 1 0 PWL(0 -2 10N -2 20N 2 2U 2 2.01U -2 4U -2 4.01U
+ -.1 6U -.1 6.0 1U .1 8U .1 8.01U -.1 10U -.1)
VDD 4 0 DC 2.5 AC 1.0
VSS 0 5 DC 2.5
CL 3 0 20P
X1 1 3 3 4 5 OPAMP . . . (Subcircuit of Table 6.6-1) . . . .DC VIN+ -2.5 2.5 0.1
.PRINT DC V(3)
.TRAN 0.05U 10U 0 10N
.PRINT TRAN V(3) V(1)
.AC DEC 10 1 10MEG
.PRINT AC VDB(3) VP(3)
.PROBE (This entry is unique to PSPICE)
.END
vin
vout
VDD
VSS
+
-3
3
1
4
5
Fig. 6.6-16A
Subckt.
-3
-2
-1
0
1
2
3
4
-3 -2 -1 0 1 2 3vO
UT
(V)
vIN(V)
ID(M5)
0
10
20
30
40
ID(M
5) μ
A
Input CMR
Fig. 240-21
Chapter 6 – Section 5 (6/24/06) Page 6.5-20
CMOS Analog Circuit Design © P.E. Allen - 2006
Example 6.6-2 - Continued
Positive PSRR of Example 6.3-1:
-20
0
20
40
60
80
10 100 1000 104 105 106 107 108
|PSR
R+(jω
)| dB
Frequency (Hz)
-100
-50
0
50
100
10 100 1000 104 105 106 107 108
Arg
[PSR
R+(jω
)] (
Deg
rees
)
Frequency (Hz)
100
Fig. 240-22
Chapter 6 – Section 5 (6/24/06) Page 6.5-21
CMOS Analog Circuit Design © P.E. Allen - 2006
Example 6.6-2 - Continued
Negative PSRR of Example 6.3-1:
10 100 1000 104 105 106 107 108
|PSR
R- (
jω)|
dB
Frequency (Hz)
-200
-150
-100
-50
0
50
100
150
200
10 100 1000 104 105 106 107 108
Arg
[PSR
R- (
jω)]
(D
egre
es)
Frequency (Hz)
20
40
60
80
100
120
Fig. 240-23
PSRR+
Chapter 6 – Section 5 (6/24/06) Page 6.5-22
CMOS Analog Circuit Design © P.E. Allen - 2006
Example 6.6-2 - Continued Large-signal and small-signal transient response of Example 6.3-1:
-1.5
-1
-0.5
0
0.5
1
1.5
0 1 2 3 4 5
Vol
ts
Time (Microseconds)
-0.15
-0.1
-0.05
0
0.05
0.1
0.15
2.5 3.0 3.5 4.0 4.5
Vol
ts
Time (Microseconds)
vin(t)
vout(t)
vin(t)
vout(t)
Fig. 240-24
M6
M7
vout
VDD
VSS
VBias-
Cc
CL
+
95μA
iCc iCL dvoutdt
Fig. 240-25
Why the negative overshoot on the slew rate? If M7 cannot sink sufficient current then the output stage slews and only responds to changes at the output via the feedback path which involves a delay.
Note that -dvout/dt -2V/0.3μs = -6.67V/μs. For a 10pF capacitor this requires 66.7μA and only 95μA-66.7μA = 28μA is available for Cc. For the positive slew rate, M6 can provide whatever current is required by the capacitors and can immediately respond to changes at the output.
Chapter 6 – Section 5 (6/24/06) Page 6.5-23
CMOS Analog Circuit Design © P.E. Allen - 2006
Example 6.6-2 - Continued
Comparison of the Simulation Results with the Specifications of Example 6.3-1:
Specification (Power supply = ±2.5V)
Design (Ex. 6.3-1)
Simulation (Ex. 1)
Open Loop Gain >5000 10,000 GB (MHz) 5 MHz 5 MHz Input CMR (Volts) -1V to 2V -1.2 V to 2.4 V, Slew Rate (V/μsec) >10 (V/μsec) +10, -7(V/μsec) Pdiss (mW) < 2mW 0.625mW
Vout range (V) ±2V +2.3V, -2.2V
PSRR+ (0) (dB) - 87
PSRR- (0) (dB) - 106 Phase margin (degrees) 60° 65° Output Resistance (k ) - 122.5k
Chapter 6 – Section 5 (6/24/06) Page 6.5-24
CMOS Analog Circuit Design © P.E. Allen - 2006
Example 6.6-3 Why is the negative-going overshoot larger than the positive-going overshoot on the small-signal transient response of the last slide? Consider the following circuit and waveform:
During the rise time, iCL = CL(dvout/dt )= 10pF(0.2V/0.1μs) = 20μA and iCc = 3pf(2V/μs) = 6μA
i6 = 95μA + 20μA + 6μA = 121μA gm6 = 1066μS (nominal was 942.5μS) During the fall time, iCL = CL(-dvout/dt) = 10pF(-0.2V/0.1μs) = -20μA and iCc = -3pf(2V/μs) = -6μA
i6 = 95μA - 20μA - 6μA = 69μA gm6 = 805μS
The dominant pole is p1 (RIgm6RIICc)-1 but the GB is gmI/Cc = 94.25μS/3pF = 31.42x106 rads/sec and stays constant. Thus we must look elsewhere for the reason. Recall that p2 gm6/CL which explains the difference.
p2(95μA) = 94.25x106 rads/sec, p2(121μA) = 106.6 x106 rads/sec, and p2(69μA) = 80.05 x106 rads/sec. Thus, the phase margin is less during the fall time than the rise time.
M6
M7
vout
VDD = 2.5V
VSS = -2.5V
CcCL
VBias
95μA
94/1i6iCL
0.1V
-0.1V
0.1μs 0.1μs
t
Fig. 240-26
iCc
Chapter 6 – Section 6 (6/24/06) Page 6.6-1
CMOS Analog Circuit Design © P.E. Allen - 2006
SECTION 6.6 - MACROMODELS FOR OP AMPS Macromodel
A macromodel is a model that captures some or all of the performance of a circuit using different components (generally simpler).
A macromodel uses resistors, capacitors, inductors, controlled sources, and some active devices (mostly diodes) to capture the essence of the performance of a complex circuit like an op amp without modeling every internal component of the op amp.
Op Amp Characterization
• Small signal, frequency independent
• Small signal, frequency dependent
• Large signal
Time independent
Time dependent
Chapter 6 – Section 6 (6/24/06) Page 6.6-2
CMOS Analog Circuit Design © P.E. Allen - 2006
SMALL SIGNAL, FREQUENCY INDEPENDENT, OP AMP MODELS Simple Model
vov1
v2
A Rid
v1
v2
Ro
voAvdRo
(v1-v2)
o
Rid
v1
v2
RoAvd (v1-v2)
1
3
1
2
3
2
4
v
(a.) (b.) (c.) Fig. 010-01 Figure 1 - (a.) Op amp symbol. (b.) Thevenin form of simple model. (c.) Norton form of simple model.
SPICE Description of Fig. 1c RID 1 2 {Rid} RO 3 0 {Ro} GAVD 0 3 1 2 {Avd/Ro}
Subcircuit SPICE Description for Fig. 1c .SUBCKT SIMPLEOPAMP 1 2 3 RID 1 2 {Rid} RO 3 0 {Ro} GAVD 0 3 1 2 {Avd/Ro} .ENDS SIMPLEOPAMP
Chapter 6 – Section 6 (6/24/06) Page 6.6-3
CMOS Analog Circuit Design © P.E. Allen - 2006
Example 6.6-1 - Use of the Simple Op Amp Model Use SPICE to find the voltage gain, vout/vin, the input resistance, Rin, and the output resistance, Rout of Fig. 2. The op amp parameters are Avd = 100,000, Rid = 1M , and Ro = 100 . Find the input resistance, Rin, the output resistance, Rout, and the voltage gain, Av, of the noninverting voltage amplifier configuration when R1 = 1k and R2 = 100k . Solution The circuit with the SPICE node numbers identified is shown in Fig. 2.
Figure 2 – Noninverting voltage amplifier for Ex. 6.6-1.
The input file for this example is given as follows. Example 1 VIN 1 0 DC 0 AC 1 XOPAMP1 1 3 2 SIMPLEOPAMP R1 3 0 1KOHM R2 2 3 100KOHM .SUBCKT SIMPLEOPAMP 1 2 3 RID 1 2 1MEGOHM RO 3 0 100OHM GAVD/RO 0 3 1 2 1000 .ENDS SIMPLEOPAMP .TF V(2) VIN .END
The command .TF finds the small signal input resistance, output resistance, and voltage or current gain of an amplifier. The results extracted from the output file are: **** SMALL-SIGNAL CHARACTERISTICS V(2)/VIN = 1.009E+02 INPUT RESISTANCE AT VIN = 9.901E+08
OUTPUT RESISTANCE AT V(2) = 1.010E-01.
+
-vin
vout
R2 = 100kΩR1 = 1kΩ
1
3
2A1
Rin Rout
Fig. 010-02
Chapter 6 – Section 6 (6/24/06) Page 6.6-4
CMOS Analog Circuit Design © P.E. Allen - 2006
Common Mode Model Electrical Model:
vo = Avd(v1-v2) + AcmRo
v1+v22
Macromodel:
Rid
1
2
Ric1
Ric2
vo
-
+3
Ro
Linear Op Amp Macromodel
Avcv12Ro
Avcv22RoAvd(v1-v2)
Ro
Fig. 010-03 Figure 3 - Simple op amp model including differential and common mode behavior. SPICE File:
.SUBCKT LINOPAMP 1 2 3 RIC1 1 0 {Ric} RID 1 2 {Rid} RIC2 2 0 {Ric}
GAVD/RO 0 3 1 2 {Avd/Ro} GAVC1/RO 0 3 1 0 {Avc/2Ro} GAVC2/RO 0 3 2 0 {Avc/2Ro} RO 3 0 {Ro} .ENDS LINOPAMP
Chapter 6 – Section 6 (6/24/06) Page 6.6-5
CMOS Analog Circuit Design © P.E. Allen - 2006
SMALL SIGNAL, FREQUENCY DEPENDENT OP AMP MODELS
Dominant Pole Model
Avd(s) = Avd(0)
(s/ 1) + 1 where 1= 1
R1C1 (dominant pole)
Model Using Passive Components:
Rid
v1
v2
vo1
2R1 C1
Avd(0)R1
(v1-v2)
3
Fig. 010-04 Figure 4 - Macromodel for the op amp including the frequency response of Avd.
Model Using Passive Components with Constant Output Resistance:
Rid
v1
v2
vo31
2R1 C1 Ro
v3Ro
4
Avd(0)R1
(v1-v2)
Fig. 010-05 Figure 5 - Frequency dependent model with constant output resistance.
Chapter 6 – Section 6 (6/24/06) Page 6.6-6
CMOS Analog Circuit Design © P.E. Allen - 2006
Example 6.6-2 - Frequency Response of the Noninverting Voltage Amplifier
Use the model of Fig. 4 to find the frequency response of Fig. 2 if the gain is +1, +10, and +100 V/V assuming that Avd(0) = 105 and 1= 100 rads/sec.
Solution The parameters of the model are R2/R1 = 0, 9, and 99. Let us additionally select Rid = 1M and Ro = 100 . We will use the circuit of Fig. 2 and insert the model as a subcircuit. The input file for this example is shown below. Example 2
VIN 1 0 DC 0 AC 1 *Unity Gain Configuration XOPAMP1 1 31 21 LINFREQOPAMP R11 31 0 15GOHM R21 21 31 1OHM *Gain of 10 Configuration XOPAMP2 1 32 22 LINFREQOPAMP
R12 32 0 1KOHM R22 22 32 9KOHM *Gain of 100 Configuration XOPAMP3 1 33 23 LINFREQOPAMP R13 33 0 1KOHM R23 23 33 99KOHM .SUBCKT LINFREQOPAMP 1 2 3 RID 1 2 1MEGOHM
GAVD/RO 0 3 1 2 1000 R1 3 0 100 C1 3 0 100UF .ENDS .AC DEC 10 100 10MEG .PRINT AC V(21) V(22) V(23) .PROBE .END
Chapter 6 – Section 6 (6/24/06) Page 6.6-7
CMOS Analog Circuit Design © P.E. Allen - 2006
Example 6.6-2 - Continued
-20dB
-10dB
0dB
10dB
20dB
30dB
40dB
100Hz 1kHz 10kHz 100kHz 1MHz 10MHz
Gain of 100
Gain of 10
Gain of 1
15.9kHz 159kHz 1.59MHz
Fig. 010-06
Figure 6 - Frequency response of the three noninverting voltage amplifiers of Ex. 6.6-2.
Chapter 6 – Section 6 (6/24/06) Page 6.6-8
CMOS Analog Circuit Design © P.E. Allen - 2006
Behavioral Frequency Model
Use of Laplace behavioral modeling capability in PSPICE. GAVD/RO 0 3 LAPLACE {V(1,2)} = {1000/(0.01s+1)}. Implements,
GAvd/Ro = Avd(s)
Ro =
Avd(0)Ro
s1 + 1
where Avd(0) = 100,000, Ro = 100 , and 1 = 100 rps
Chapter 6 – Section 6 (6/24/06) Page 6.6-9
CMOS Analog Circuit Design © P.E. Allen - 2006
Differential and Common Mode Frequency Dependent Models
Figure 7 - Op amp macromodel for separate differential and common voltage gain frequency responses.
Rid
2
Ric1
Ric2
vo
-
+3
Op Amp Macromodel
Avcv12Ro
Avcv22Ro
Avd(v1-v2)Ro
v3Ro
R1
R2C2
C1
4
5
Rov4Ro
1
Fig. 010-07
Chapter 6 – Section 6 (6/24/06) Page 6.6-10
CMOS Analog Circuit Design © P.E. Allen - 2006
Zeros in the Transfer Function
Models: 3
Avd(v1-v2)R1
R1C1 vo
-
+
v3Ro
RokAvdRo
(v1-v2)
4
(a.) (b.)
3
Avd(v1-v2)Ro
Ro
L1
vo
-
+
4
Fig. 010-08 Figure 8 - (a.) Independent zero model. (b.) Method of modeling zeros without introducing new nodes. Inductor:
Vo(s) = Avd(0)
Ro (sL1 + Ro) [V1(s)-V2(s)] = Avd(0)s
Ro/L1 + 1 [V1(s)-V2(s)] .
Feedforward:
Vo(s) = Avd(0)
(s/ 1) +1 [ ]1+k(s/ 1)+k [V1(s)-V2(s)] .
The zero can be expressed as z1 = - 1 1 + 1k
where k can be + or - by reversing the direction of the current source.
Chapter 6 – Section 6 (6/24/06) Page 6.6-11
CMOS Analog Circuit Design © P.E. Allen - 2006
Example 6.6-3 - Modeling Zeros in the Op Amp Frequency Response
Use the technique of Fig. 8b to model an op amp with a differential voltage gain of 100,000, a pole at 100rps, an output resistance of 100 , and a zero in the right-half, complex frequency plane at 107 rps. Solution The transfer function we want to model is given as
Vo(s) = 105(s/107 - 1)
(s/100 + 1) .
Let us arbitrarily select R1 as 100k which will make the GAVD/R1 gain unity. To get the pole at 100rps, C1 = 1/(100R1) = 0.1μF. Next, we want z1 to be 107 rps. Since 1 = 100rps, then Eq. (6) gives k as -10-5. The following input file verifies this model. Example 3
VIN 1 0 DC 0 AC 1 XOPAMP1 1 0 2 LINFREQOPAMP .SUBCKT LINFREQOPAMP 1 2 4 RID 1 2 1MEGOHM GAVD/R1 0 3 1 2 1 R1 3 0 100KOHM C1 3 0 0.1UF
GV3/RO 0 4 3 0 0.01 GAVD/RO 4 0 1 2 0.01 RO 4 0 100 .ENDS .AC DEC 10 1 100MEG .PRINT AC V(2) VDB(2) VP(2) .PROBE .END
Chapter 6 – Section 6 (6/24/06) Page 6.6-12
CMOS Analog Circuit Design © P.E. Allen - 2006
Example 6.6-3 - Continued The asymptotic magnitude frequency response of this simulation is shown in Fig. 9. We note that although the frequency response is plotted in Hertz, there is a pole at 100rps (15.9Hz) and a zero at 1.59MHz (10Mrps). Unless we examined the phase shift, it is not possible to determine whether the zero is in the RHP or LHP of the complex frequency axis.
VD
B(2
)
0dB
20dB
40dB
60dB
80dB
100dB
1Hz 100Hz10Hz 1kHz 10kHz 100kHz 1MHz 10MHz
15.9Hz or 100rps
1.59MHz or 10Mrps
Frequency Fig. 010-09 Figure 9 - Asymptotic magnitude frequency response of the op amp model of Ex. 6.6-3.
Chapter 6 – Section 6 (6/24/06) Page 6.6-13
CMOS Analog Circuit Design © P.E. Allen - 2006
LARGE SIGNAL MACROMODELS FOR THE OP AMP Output and Input Voltage Limitations
Figure 10 - Op amp macromodel that limits the input and output voltages.
Subcircuit Description .SUBCKT NONLINOPAMP 1 2 3 RIC1 1 0 {Ricm} RLIM1 1 4 0.1 D1 4 6 IDEALMOD VIH1 6 0 {VIH1} D2 7 4 IDEALMOD VIL1 7 0 {VIL1} RID 4 5 {Rid} RIC2 2 0 {Ricm} RLIM2 2 5 0.1
D3 5 8 IDEALMOD VIH2 8 0 {VIH1}
D4 9 5 IDEALMOD VIL2 9 0 {VIL2} GAVD/RO 0 3 4 5 {Avd/Ro} GAVC1/RO 0 3 4 0 {Avc/Ro} GAVC2/RO 0 3 5 0 {Avc/Ro} RO 3 0 {Ro} D5 3 10 IDEALMOD VOH 10 0 {VOH} D6 11 3 IDEALMOD VOL 11 0 {VOL} .MODEL IDEALMOD D N=0.001
.ENDS
Nonlinear Op Amp Macromodel
10 vo
-
+3
Ro
Avcv42Ro 11
D6+
-
+
-
D5
VOH VOL
Avcv52Ro
Rid
1
2 5
Ric1
Ric2
7D1 D2
+
-
+
-
RLIM 4
6
D3 D49+
-
+
-
RLIM
8
VIH1
VIH2 VIL2
VIL1
AvdRo
(v4-v5)
Fig. 010-10
Chapter 6 – Section 6 (6/24/06) Page 6.6-14
CMOS Analog Circuit Design © P.E. Allen - 2006
Example 6.6-4 - Illustration of the Voltage Limits of the Op Amp
Use the macromodel of Fig. 10 to plot vOUT as a function of vIN for the noninverting, unity gain, voltage amplifier when vIN is varied from -15V to +15V. The op amp parameters are Avd(0) = 100,000, Rid = 1M , Ricm = 100M , Avc(0) = 10, Ro = 100 , VOH = -VOL = 10V, VIH1 =VIH2 = -VIL1 = -VIL2 = 5V. Solution The input file for this example is given below. Example 4
VIN 1 0 DC 0 XOPAMP 1 2 2 NONLINOPAMP .SUBCKT NONLINOPAMP 1 2 3 RIC1 1 0 100MEG RLIM1 1 4 0.1 D1 4 6 IDEALMOD VIH1 6 0 5V D2 7 4 IDEALMOD
VIL1 7 0 -5V RID 4 5 1MEG RIC2 2 0 100MEG RLIM2 2 5 0.1 D3 5 8 IDEALMOD VIH2 8 0 5V D4 9 5 IDEALMOD VIL2 9 0 -5v GAVD/RO 0 3 4 5 1000 GAVC1/2RO 0 3 4 0 0.05 GAVC2/2RO 0 3 5 0 0.05
RO 3 0 100 D5 3 10 IDEALMOD VOH 10 0 10V D6 11 3 IDEALMOD VOL 11 0 -10V .MODEL IDEALMOD D N=0.0001 .ENDS .DC VIN -15 15 0.1 .PRINT V(2) .PROBE .END
Chapter 6 – Section 6 (6/24/06) Page 6.6-15
CMOS Analog Circuit Design © P.E. Allen - 2006
Example 6.6-4 - Illustration of the Voltage Limits of the Op Amp - Continued
-7.5V
-5V
-2.5V
0V
2.5V
5V
7.5V
-10V -5V 0V 5V 10V
V(2)
VIN Fig. 010-11 Figure 11 - Simulation results for Ex. 6.6-4.
Chapter 6 – Section 6 (6/24/06) Page 6.6-16
CMOS Analog Circuit Design © P.E. Allen - 2006
Output Current Limiting Technique:
D1
D2
D3
D4
ILimit
Io Io
Io2
ILimit2
Io2
Io2
Io2
ILimit2
Fig. 010-12 Macromodel for Output Voltage and Current Limiting:
Rid
v1
v2
Ro
vo31
2AvdRo
(v1-v2)7+
-
8+
-
D1
D2
D3
D4D5 D6
ILimit VOH VOL
45
6
Fig. 010-13
Chapter 6 – Section 6 (6/24/06) Page 6.6-17
CMOS Analog Circuit Design © P.E. Allen - 2006
Example 6.6-5 - Influence of Current Limiting on the Amplifier Voltage Transfer Curve Use the model above to illustrate the influence of current limiting on the voltage transfer curve of an inverting gain of one amplifier. Assume the VOH = -VOL = 10V, VIH = -VIL = 10V, the maximum output current is ±20mA, and R1 = R2 = RL = 500 where RL is a resistor connected from the output to ground. Otherwise, the op amp is ideal. Solution For the ideal op amp we will choose Avd = 100,000, Rid = 1M , and Ro = 100 and assume one cannot tell the difference between these parameters and the ideal parameters. The remaining model parameters are VOH = -VOL = 10V and ILimit = ±20mA. Example 5 - Influence of Current Limiting on the Amplifier Voltage Transfer Curve
VIN 1 0 DC 0 R1 1 2 500 R2 2 3 500 RL 3 0 500 XOPAMP 0 2 3 NONLINOPAMP .SUBCKT NONLINOPAMP 1 2 3 RID 1 2 1MEGOHM GAVD 0 4 1 2 1000 RO 4 0 100 D1 3 5 IDEALMOD D2 6 3 IDEALMOD D3 4 5 IDEALMOD
D4 6 4 IDEALMOD ILIMIT 5 6 20MA D5 3 7 IDEALMOD VOH 7 0 10V D6 8 3 IDEALMOD VOL 8 0 -10V .MODEL IDEALMOD D N=0.00001 .ENDS .DC VIN -15 15 0.1 .PRINT DC V(3) .PROBE .END
Chapter 6 – Section 6 (6/24/06) Page 6.6-18
CMOS Analog Circuit Design © P.E. Allen - 2006
Example 6.6-5 - Continued
The resulting plot of the output voltage, v3, as a function of the input voltage, vIN is shown in Fig. 14.
-10V
-5V
0V
5V
10V
V(3
)
-15V -10V -5V 0V 5V 10V 15VVIN Fig. 010-14
Figure 14 - Results of Example 6.6-5.
Chapter 6 – Section 6 (6/24/06) Page 6.6-19
CMOS Analog Circuit Design © P.E. Allen - 2006
Slew Rate Limiting (Time Dependency)
Slew Rate:
dvodt =
±ISRC1
= Slew Rate
Macromodel:
Rid
v1
v2
vo3
1
2R1
C1Avd(0)R1
(v1-v2)
D1
D2
D3
D4
6
7Ro
v4-v5Ro
45
ISR
Fig. 010-15
Chapter 6 – Section 6 (6/24/06) Page 6.6-20
CMOS Analog Circuit Design © P.E. Allen - 2006
Example 6.6-6 - Simulation of the Slew Rate of A Noninverting Voltage Amplifier
Let the gain of a noninverting voltage amplifier be 1. If the input signal is given as vin(t) = 10 sin(4x105 t) use the computer to find the output voltage if the slew rate of the op amp is 10V/μs.
Solution We can calculate that the op amp should slew when the frequency is 159kHz. Let us assume the op amp parameters of Avd = 100,000, 1 = 100rps, Rid = 1M , and Ro = 100 . The simulation input file based on the macromodel of Fig. 15 is given below. Example 6.6-6 - Simulation of slew rate limitation
VIN 1 0 SIN(0 10 200K) XOPAMP 1 2 2 NONLINOPAMP .SUBCKT NONLINOPAMP 1 2 3 RID 1 2 1MEGOHM GAVD/R1 0 4 1 2 1 R1 4 0 100KOHM C1 4 5 0.1UF D1 0 6 IDEALMOD
D2 7 0 IDEALMOD D3 5 6 IDEALMOD D4 7 5 IDEALMOD ISR 6 7 1A GVO/R0 0 3 4 5 0.01 RO 3 0 100 .MODEL IDEALMOD D N=0.0001 .ENDS
Chapter 6 – Section 6 (6/24/06) Page 6.6-21
CMOS Analog Circuit Design © P.E. Allen - 2006
Example 6.6-6 - Continued The simulation results are shown in Fig. 16. The input waveform is shown along with the output waveform. The influence of the slew rate causes the output waveform not to be equal to the input waveform.
-10V
-5V
0V
5V
10V
0μs 2μs 4μs 6μs 8μs 10μs
InputVoltage
OutputVoltage
Time Fig. 010-16 Figure 16 - Results of Ex. 6.6-6 on modeling the slew rate of an op amp.
Chapter 6 – Section 6 (6/24/06) Page 6.6-22
CMOS Analog Circuit Design © P.E. Allen - 2006
SPICE Op Amp Library Models
Macromodels developed from the data sheet for various components.
Key Aspects of Op Amp Macromodels:
• Use the simplest op amp macromodel for a given simulation.
• All things being equal, use the macromodel with the min. no. of nodes.
• Use the SUBCKT feature for repeated use of the macromodel.
• Be sure to verify the correctness of the macromodels before using.
• Macromodels are a good means of trading simulation completeness for decreased simulation time.
Chapter 6 – Section 7 (6/24/06) Page 6.7-1
CMOS Analog Circuit Design © P.E. Allen - 2006
SECTION 6.8 - SUMMARY Topics
• Design of CMOS op amps • Compensation of op amps - Miller - Self-compensating - Feedforward • Two-stage op amp design • Cascode op amps • Simulation and measurement of op amps • Macromodels of op amps
Chapter Objective
Purpose of this chapter is to introduce the simple two-stage op amp and the cascode op amp to illustrate the concepts of op amp design and to form the starting point for the improvement of performance of the next chapter.
Design Procedures
The design procedures given in this chapter are for the purposes of understanding and applying the design relationships and should not be followed rigorously as the designer gains experience.